在 Android EditText 中限制小數位數

Question

我正在嘗試編寫一個可以幫助您管理財務的應用程序。 我正在使用EditText字段，用戶可以在其中指定金額。

我將inputType設置為numberDecimal ，這可以正常工作，除了這允許人們輸入諸如123.122類的數字，這對於金錢來說並不完美。

有沒有辦法將小數點后的字符數限制為兩個？

Answer 1

更優雅的方法是使用正則表達式 (regex)，如下所示：

public class DecimalDigitsInputFilter implements InputFilter {

Pattern mPattern;

public DecimalDigitsInputFilter(int digitsBeforeZero,int digitsAfterZero) {
    mPattern=Pattern.compile("[0-9]{0," + (digitsBeforeZero-1) + "}+((\\.[0-9]{0," + (digitsAfterZero-1) + "})?)||(\\.)?");
}

@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {

        Matcher matcher=mPattern.matcher(dest);       
        if(!matcher.matches())
            return "";
        return null;
    }

}

要使用它，請執行以下操作：

editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});

Answer 2

不使用正則表達式的更簡單的解決方案：

import android.text.InputFilter;
import android.text.Spanned;

/**
 * Input filter that limits the number of decimal digits that are allowed to be
 * entered.
 */
public class DecimalDigitsInputFilter implements InputFilter {

  private final int decimalDigits;

  /**
   * Constructor.
   * 
   * @param decimalDigits maximum decimal digits
   */
  public DecimalDigitsInputFilter(int decimalDigits) {
    this.decimalDigits = decimalDigits;
  }

  @Override
  public CharSequence filter(CharSequence source,
      int start,
      int end,
      Spanned dest,
      int dstart,
      int dend) {


    int dotPos = -1;
    int len = dest.length();
    for (int i = 0; i < len; i++) {
      char c = dest.charAt(i);
      if (c == '.' || c == ',') {
        dotPos = i;
        break;
      }
    }
    if (dotPos >= 0) {

      // protects against many dots
      if (source.equals(".") || source.equals(","))
      {
          return "";
      }
      // if the text is entered before the dot
      if (dend <= dotPos) {
        return null;
      }
      if (len - dotPos > decimalDigits) {
        return "";
      }
    }

    return null;
  }

}

使用：

editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(2)});

Answer 3

InputFilter這個實現解決了這個問題。

import android.text.SpannableStringBuilder;
import android.text.Spanned;
import android.text.method.DigitsKeyListener;

public class MoneyValueFilter extends DigitsKeyListener {
    public MoneyValueFilter() {
        super(false, true);
    }

    private int digits = 2;

    public void setDigits(int d) {
        digits = d;
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end,
            Spanned dest, int dstart, int dend) {
        CharSequence out = super.filter(source, start, end, dest, dstart, dend);

        // if changed, replace the source
        if (out != null) {
            source = out;
            start = 0;
            end = out.length();
        }

        int len = end - start;

        // if deleting, source is empty
        // and deleting can't break anything
        if (len == 0) {
            return source;
        }

        int dlen = dest.length();

        // Find the position of the decimal .
        for (int i = 0; i < dstart; i++) {
            if (dest.charAt(i) == '.') {
                // being here means, that a number has
                // been inserted after the dot
                // check if the amount of digits is right
                return (dlen-(i+1) + len > digits) ? 
                    "" :
                    new SpannableStringBuilder(source, start, end);
            }
        }

        for (int i = start; i < end; ++i) {
            if (source.charAt(i) == '.') {
                // being here means, dot has been inserted
                // check if the amount of digits is right
                if ((dlen-dend) + (end-(i + 1)) > digits)
                    return "";
                else
                    break;  // return new SpannableStringBuilder(source, start, end);
            }
        }

        // if the dot is after the inserted part,
        // nothing can break
        return new SpannableStringBuilder(source, start, end);
    }
}

Answer 4

這是一個示例InputFilter ，它只允許小數點前最多 4 位數字和小數點后最多 1 位數字。

edittext 允許的值： 555.2 、 555 、 .2

編輯文本塊的值： 55555.2 、 055.2 、 555.42

        InputFilter filter = new InputFilter() {
        final int maxDigitsBeforeDecimalPoint=4;
        final int maxDigitsAfterDecimalPoint=1;

        @Override
        public CharSequence filter(CharSequence source, int start, int end,
                Spanned dest, int dstart, int dend) {
                StringBuilder builder = new StringBuilder(dest);
                builder.replace(dstart, dend, source
                        .subSequence(start, end).toString());
                if (!builder.toString().matches(
                        "(([1-9]{1})([0-9]{0,"+(maxDigitsBeforeDecimalPoint-1)+"})?)?(\\.[0-9]{0,"+maxDigitsAfterDecimalPoint+"})?"

                        )) {
                    if(source.length()==0)
                        return dest.subSequence(dstart, dend);
                    return "";
                }

            return null;

        }
    };

    mEdittext.setFilters(new InputFilter[] { filter });

Answer 5

我為@Pinhassi 解決方案做了一些修復。 它處理一些情況：

1.你可以將光標移動到任何地方

2.減號處理

3.digitsbefore = 2 和digitsafter = 4 你輸入12.4545。 那么如果你想去掉“.”，就不允許了。

public class DecimalDigitsInputFilter implements InputFilter {
    private int mDigitsBeforeZero;
    private int mDigitsAfterZero;
    private Pattern mPattern;

    private static final int DIGITS_BEFORE_ZERO_DEFAULT = 100;
    private static final int DIGITS_AFTER_ZERO_DEFAULT = 100;

    public DecimalDigitsInputFilter(Integer digitsBeforeZero, Integer digitsAfterZero) {
    this.mDigitsBeforeZero = (digitsBeforeZero != null ? digitsBeforeZero : DIGITS_BEFORE_ZERO_DEFAULT);
    this.mDigitsAfterZero = (digitsAfterZero != null ? digitsAfterZero : DIGITS_AFTER_ZERO_DEFAULT);
    mPattern = Pattern.compile("-?[0-9]{0," + (mDigitsBeforeZero) + "}+((\\.[0-9]{0," + (mDigitsAfterZero)
        + "})?)||(\\.)?");
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
    String replacement = source.subSequence(start, end).toString();
    String newVal = dest.subSequence(0, dstart).toString() + replacement
        + dest.subSequence(dend, dest.length()).toString();
    Matcher matcher = mPattern.matcher(newVal);
    if (matcher.matches())
        return null;

    if (TextUtils.isEmpty(source))
        return dest.subSequence(dstart, dend);
    else
        return "";
    }
}

Answer 6

我不喜歡其他解決方案，我創建了自己的解決方案。 使用此解決方案，您不能在點前輸入超過 MAX_BEFORE_POINT 的數字，並且小數點不能超過 MAX_DECIMAL。

您不能輸入過多的數字，沒有其他影響！ 另外如果你寫“。” 它鍵入“0”。

1. 將布局中的 EditText 設置為：

   android:inputType="numberDecimal"

2. 在您的 onCreate 中添加偵聽器。 如果要修改點之前和之后的位數，請編輯對 PerfectDecimal(str, NUMBER_BEFORE_POINT, NUMBER_DECIMALS) 的調用，這里設置為 3 和 2

    EditText targetEditText = (EditText)findViewById(R.id.targetEditTextLayoutId); targetEditText.addTextChangedListener(new TextWatcher() { public void onTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {} public void beforeTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {} public void afterTextChanged(Editable arg0) { String str = targetEditText.getText().toString(); if (str.isEmpty()) return; String str2 = PerfectDecimal(str, 3, 2); if (!str2.equals(str)) { targetEditText.setText(str2); targetEditText.setSelection(str2.length()); } } });

3. 包括這個功能：

    public String PerfectDecimal(String str, int MAX_BEFORE_POINT, int MAX_DECIMAL){ if(str.charAt(0) == '.') str = "0"+str; int max = str.length(); String rFinal = ""; boolean after = false; int i = 0, up = 0, decimal = 0; char t; while(i < max){ t = str.charAt(i); if(t != '.' && after == false){ up++; if(up > MAX_BEFORE_POINT) return rFinal; }else if(t == '.'){ after = true; }else{ decimal++; if(decimal > MAX_DECIMAL) return rFinal; } rFinal = rFinal + t; i++; }return rFinal; }

它完成了！

Answer 7

我通過以下方式在TextWatcher的幫助下實現了這一點

final EditText et = (EditText) findViewById(R.id.EditText1);
int count = -1;
et.addTextChangedListener(new TextWatcher() {
    public void onTextChanged(CharSequence arg0, int arg1, int arg2,int arg3) {             

    }
    public void beforeTextChanged(CharSequence arg0, int arg1,int arg2, int arg3) {             

    }

    public void afterTextChanged(Editable arg0) {
        if (arg0.length() > 0) {
            String str = et.getText().toString();
            et.setOnKeyListener(new OnKeyListener() {
                public boolean onKey(View v, int keyCode, KeyEvent event) {
                    if (keyCode == KeyEvent.KEYCODE_DEL) {
                        count--;
                        InputFilter[] fArray = new InputFilter[1];
                        fArray[0] = new InputFilter.LengthFilter(100);
                        et.setFilters(fArray);
                        //change the edittext's maximum length to 100. 
                        //If we didn't change this the edittext's maximum length will
                        //be number of digits we previously entered.
                    }
                    return false;
                }
            });
            char t = str.charAt(arg0.length() - 1);
            if (t == '.') {
                count = 0;
            }
            if (count >= 0) {
                if (count == 2) {                        
                    InputFilter[] fArray = new InputFilter[1];
                    fArray[0] = new InputFilter.LengthFilter(arg0.length());
                    et.setFilters(fArray);
                    //prevent the edittext from accessing digits 
                    //by setting maximum length as total number of digits we typed till now.
                }
                count++;
            }
        }
    }
});

該解決方案不允許用戶在小數點后輸入超過兩位數。 您也可以在小數點前輸入任意數量的數字。 我希望這將有所幫助。 謝謝你。

Answer 8

要求是小數點后2位。 小數點前的位數應該沒有限制。 所以，解決方案應該是，

public class DecimalDigitsInputFilter implements InputFilter {

    Pattern mPattern;

    public DecimalDigitsInputFilter() {
        mPattern = Pattern.compile("[0-9]*+((\\.[0-9]?)?)||(\\.)?");
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
        Matcher matcher = mPattern.matcher(dest);
        if (!matcher.matches())
            return "";
        return null;
    }
}

並將其用作，

mEditText.setFilters(new InputFilter[]{new DecimalDigitsInputFilter()});

感謝@Pinhassi 的啟發。

Answer 9

我提出的 InputFilter 允許您配置小數位前后的位數。 此外，它不允許前導零。

public class DecimalDigitsInputFilter implements InputFilter
{
    Pattern pattern;

    public DecimalDigitsInputFilter(int digitsBeforeDecimal, int digitsAfterDecimal)
    {
        pattern = Pattern.compile("(([1-9]{1}[0-9]{0," + (digitsBeforeDecimal - 1) + "})?||[0]{1})((\\.[0-9]{0," + digitsAfterDecimal + "})?)||(\\.)?");
    }

    @Override public CharSequence filter(CharSequence source, int sourceStart, int sourceEnd, Spanned destination, int destinationStart, int destinationEnd)
    {
        // Remove the string out of destination that is to be replaced.
        String newString = destination.toString().substring(0, destinationStart) + destination.toString().substring(destinationEnd, destination.toString().length());

        // Add the new string in.
        newString = newString.substring(0, destinationStart) + source.toString() + newString.substring(destinationStart, newString.length());

        // Now check if the new string is valid.
        Matcher matcher = pattern.matcher(newString);

        if(matcher.matches())
        {
            // Returning null indicates that the input is valid.
            return null;
        }

        // Returning the empty string indicates the input is invalid.
        return "";
    }
}

// To use this InputFilter, attach it to your EditText like so:
final EditText editText = (EditText) findViewById(R.id.editText);

EditText.setFilters(new InputFilter[]{new DecimalDigitsInputFilter(4, 4)});

Answer 10

我的解決方案很簡單，而且效果很好！

public class DecimalInputTextWatcher implements TextWatcher {

private String mPreviousValue;
private int mCursorPosition;
private boolean mRestoringPreviousValueFlag;
private int mDigitsAfterZero;
private EditText mEditText;

public DecimalInputTextWatcher(EditText editText, int digitsAfterZero) {
    mDigitsAfterZero = digitsAfterZero;
    mEditText = editText;
    mPreviousValue = "";
    mRestoringPreviousValueFlag = false;
}

@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
    if (!mRestoringPreviousValueFlag) {
        mPreviousValue = s.toString();
        mCursorPosition = mEditText.getSelectionStart();
    }
}

@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
}

@Override
public void afterTextChanged(Editable s) {
    if (!mRestoringPreviousValueFlag) {

        if (!isValid(s.toString())) {
            mRestoringPreviousValueFlag = true;
            restorePreviousValue();
        }

    } else {
        mRestoringPreviousValueFlag = false;
    }
}

private void restorePreviousValue() {
    mEditText.setText(mPreviousValue);
    mEditText.setSelection(mCursorPosition);
}

private boolean isValid(String s) {
    Pattern patternWithDot = Pattern.compile("[0-9]*((\\.[0-9]{0," + mDigitsAfterZero + "})?)||(\\.)?");
    Pattern patternWithComma = Pattern.compile("[0-9]*((,[0-9]{0," + mDigitsAfterZero + "})?)||(,)?");

    Matcher matcherDot = patternWithDot.matcher(s);
    Matcher matcherComa = patternWithComma.matcher(s);

    return matcherDot.matches() || matcherComa.matches();
}
}

用法：

myTextEdit.addTextChangedListener(new DecimalInputTextWatcher(myTextEdit, 2));

Answer 11

實現這一目標的最簡單方法是：

et.addTextChangedListener(new TextWatcher() {
    public void onTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {
        String text = arg0.toString();
        if (text.contains(".") && text.substring(text.indexOf(".") + 1).length() > 2) {
            et.setText(text.substring(0, text.length() - 1));
            et.setSelection(et.getText().length());
        }
    }

    public void beforeTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {

    }

    public void afterTextChanged(Editable arg0) {
    }
});

Answer 12

稍微改進了@Pinhassi 解決方案。

效果很好。 它驗證連接的字符串。

public class DecimalDigitsInputFilter implements InputFilter {

Pattern mPattern;

public DecimalDigitsInputFilter() {
    mPattern = Pattern.compile("([1-9]{1}[0-9]{0,2}([0-9]{3})*(\\.[0-9]{0,2})?|[1-9]{1}[0-9]{0,}(\\.[0-9]{0,2})?|0(\\.[0-9]{0,2})?|(\\.[0-9]{1,2})?)");

}

@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {

    String formatedSource = source.subSequence(start, end).toString();

    String destPrefix = dest.subSequence(0, dstart).toString();

    String destSuffix = dest.subSequence(dend, dest.length()).toString();

    String result = destPrefix + formatedSource + destSuffix;

    result = result.replace(",", ".");

    Matcher matcher = mPattern.matcher(result);

    if (matcher.matches()) {
        return null;
    }

    return "";
}

 }

Answer 13

我修改了上述解決方案並創建了以下解決方案。 您可以設置小數點前后的位數。

public class DecimalDigitsInputFilter implements InputFilter {

private final Pattern mPattern;

public DecimalDigitsInputFilter(int digitsBeforeZero, int digitsAfterZero) {
    mPattern = Pattern.compile(String.format("[0-9]{0,%d}(\\.[0-9]{0,%d})?", digitsBeforeZero, digitsAfterZero));
}

@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
    Matcher matcher = mPattern.matcher(createResultString(source, start, end, dest, dstart, dend));
    if (!matcher.matches())
        return "";
    return null;
}

private String createResultString(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
    String sourceString = source.toString();
    String destString = dest.toString();
    return destString.substring(0, dstart) + sourceString.substring(start, end) + destString.substring(dend);
}

}

Answer 14

在將字符串放入 TextView 之前，請嘗試使用NumberFormat.getCurrencyInstance() 對其進行格式化。

就像是：

NumberFormat currency = NumberFormat.getCurrencyInstance();
myTextView.setText(currency.format(dollars));

編輯- 我在文檔中找不到貨幣的 inputType。 我想這是因為有些貨幣不遵循相同的小數位規則，例如日元。

正如 LeffelMania 所提到的，您可以通過將上述代碼與在EditText上設置的TextWatcher一起使用來更正用戶輸入。

Answer 15

DecimalFormat form = new DecimalFormat("#.##", new DecimalFormatSymbols(Locale.US));
    EditText et; 
    et.setOnEditorActionListener(new TextView.OnEditorActionListener() {
        @Override
        public boolean onEditorAction(TextView v, int actionId, KeyEvent event) {

        if (actionId == EditorInfo.IME_ACTION_DONE) {
            double a = Double.parseDouble(et.getText().toString());
            et.setText(form.format(a));
        }
        return false;
    }
});

這樣做是當您退出編輯階段時，它將字段格式化為正確的格式。 目前，它只有 2 個十進制字符。 我認為這是很容易做到這一點的方法。

Answer 16

這里的所有答案都非常復雜，我試圖讓它變得更簡單。看看我的代碼並自己決定 -

int temp  = 0;
int check = 0;

editText.addTextChangedListener(new TextWatcher() {

    @Override
    public void onTextChanged(CharSequence s, int start, int before, int count) {

        if(editText.getText().toString().length()<temp)
        {
            if(!editText.getText().toString().contains("."))
                editText.setFilters(new InputFilter[] { new InputFilter.LengthFilter(editText.getText().toString().length()-1) });
            else
                editText.setFilters(new InputFilter[] { new InputFilter.LengthFilter(editText.getText().toString().length()+1) });

        }

        if(!editText.getText().toString().contains("."))
        {
            editText.setFilters(new InputFilter[] { new InputFilter.LengthFilter(editText.getText().toString().length()+1) });
            check=0;
        }


        else if(check==0)
        {
            check=1;
            editText.setFilters(new InputFilter[] { new InputFilter.LengthFilter(editText.getText().toString().length()+2) });
        }
    }

    @Override
    public void beforeTextChanged(CharSequence s, int start, int count,
            int after) {
        temp = editText.getText().toString().length();


    }

    @Override
    public void afterTextChanged(Editable s) {
        // TODO Auto-generated method stub

    }
});

Answer 17

我真的很喜歡 Pinhassi 的回答，但注意到在用戶輸入小數點后指定的數字后，您無法再在小數點左側輸入文本。 問題是該解決方案只測試之前輸入的文本，而不是當前輸入的文本。 所以這是我的解決方案，它將新字符插入到原始文本中進行驗證。

package com.test.test;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

import android.text.InputFilter;
import android.text.Spanned;
import android.util.Log;

public class InputFilterCurrency implements InputFilter {
    Pattern moPattern;

    public InputFilterCurrency(int aiMinorUnits) {
        // http://www.regexplanet.com/advanced/java/index.html
        moPattern=Pattern.compile("[0-9]*+((\\.[0-9]{0,"+ aiMinorUnits + "})?)||(\\.)?");

    } // InputFilterCurrency

    @Override
    public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
        String lsStart  = "";
        String lsInsert = "";
        String lsEnd    = "";
        String lsText   = "";

        Log.d("debug", moPattern.toString());
        Log.d("debug", "source: " + source + ", start: " + start + ", end:" + end + ", dest: " + dest + ", dstart: " + dstart + ", dend: " + dend );

        lsText = dest.toString();

        // If the length is greater then 0, then insert the new character
        // into the original text for validation
        if (lsText.length() > 0) {

            lsStart = lsText.substring(0, dstart);
            Log.d("debug", "lsStart : " + lsStart);
            // Check to see if they have deleted a character
            if (source != "") {
                lsInsert = source.toString();
                Log.d("debug", "lsInsert: " + lsInsert);
            } // if
            lsEnd = lsText.substring(dend);
            Log.d("debug", "lsEnd   : " + lsEnd);
            lsText = lsStart + lsInsert + lsEnd;
            Log.d("debug", "lsText  : " + lsText);

        } // if

        Matcher loMatcher = moPattern.matcher(lsText);
        Log.d("debug", "loMatcher.matches(): " + loMatcher.matches() + ", lsText: " + lsText);
        if(!loMatcher.matches()) {
            return "";
        }
        return null;

    } // CharSequence

} // InputFilterCurrency

以及設置 editText 過濾器的調用

editText.setFilters(new InputFilter[] {new InputFilterCurrency(2)});

Ouput with two decimal places
05-22 15:25:33.434: D/debug(30524): [0-9]*+((\.[0-9]{0,2})?)||(\.)?
05-22 15:25:33.434: D/debug(30524): source: 5, start: 0, end:1, dest: 123.4, dstart: 5, dend: 5
05-22 15:25:33.434: D/debug(30524): lsStart : 123.4
05-22 15:25:33.434: D/debug(30524): lsInsert: 5
05-22 15:25:33.434: D/debug(30524): lsEnd   : 
05-22 15:25:33.434: D/debug(30524): lsText  : 123.45
05-22 15:25:33.434: D/debug(30524): loMatcher.matches(): true, lsText: 123.45

Ouput inserting a 5 in the middle
05-22 15:26:17.624: D/debug(30524): [0-9]*+((\.[0-9]{0,2})?)||(\.)?
05-22 15:26:17.624: D/debug(30524): source: 5, start: 0, end:1, dest: 123.45, dstart: 2, dend: 2
05-22 15:26:17.624: D/debug(30524): lsStart : 12
05-22 15:26:17.624: D/debug(30524): lsInsert: 5
05-22 15:26:17.624: D/debug(30524): lsEnd   : 3.45
05-22 15:26:17.624: D/debug(30524): lsText  : 1253.45
05-22 15:26:17.624: D/debug(30524): loMatcher.matches(): true, lsText: 1253.45

Answer 18

我改進了 Pinhassi 使用正則表達式的解決方案，因此它也可以正確處理邊緣情況。 在檢查輸入是否正確之前，首先按照 android 文檔的描述構造最終的字符串。

public class DecimalDigitsInputFilter implements InputFilter {

    private Pattern mPattern;

    private static final Pattern mFormatPattern = Pattern.compile("\\d+\\.\\d+");

    public DecimalDigitsInputFilter(int digitsBeforeDecimal, int digitsAfterDecimal) {
        mPattern = Pattern.compile(
            "^\\d{0," + digitsBeforeDecimal + "}([\\.,](\\d{0," + digitsAfterDecimal +
                "})?)?$");
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end, Spanned dest, 
                               int dstart, int dend) {

        String newString =
            dest.toString().substring(0, dstart) + source.toString().substring(start, end) 
            + dest.toString().substring(dend, dest.toString().length());

        Matcher matcher = mPattern.matcher(newString);
        if (!matcher.matches()) {
            return "";
        }
        return null;
    }
}

用法：

editText.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});

Answer 19

Simple Helper 類是為了防止用戶在小數點后輸入超過 2 位數字：

public class CostFormatter  implements TextWatcher {

private final EditText costEditText;

public CostFormatter(EditText costEditText) {
    this.costEditText = costEditText;
}

@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
}

@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
}

@Override
public synchronized void afterTextChanged(final Editable text) {
    String cost = text.toString().trim();

    if(!cost.endsWith(".") && cost.contains(".")){
        String numberBeforeDecimal = cost.split("\\.")[0];
        String numberAfterDecimal = cost.split("\\.")[1];

        if(numberAfterDecimal.length() > 2){
            numberAfterDecimal = numberAfterDecimal.substring(0, 2);
        }
        cost = numberBeforeDecimal + "." + numberAfterDecimal;
    }
    costEditText.removeTextChangedListener(this);
    costEditText.setText(cost);
    costEditText.setSelection(costEditText.getText().toString().trim().length());
    costEditText.addTextChangedListener(this);
}
}

Answer 20

我已經更改了答案 №6（由 Favas Kv），因為在那里您可以將點放在第一個位置。

final InputFilter [] filter = { new InputFilter() {

    @Override
    public CharSequence filter(CharSequence source, int start, int end,
                               Spanned dest, int dstart, int dend) {
        StringBuilder builder = new StringBuilder(dest);
        builder.replace(dstart, dend, source
                .subSequence(start, end).toString());
        if (!builder.toString().matches(
                "(([1-9]{1})([0-9]{0,4})?(\\.)?)?([0-9]{0,2})?"

        )) {
            if(source.length()==0)
                return dest.subSequence(dstart, dend);
            return "";
        }
        return null;
    }
}};

Answer 21

就像其他人說的，我在我的項目中添加了這個類並將過濾器設置為我想要的EditText 。

過濾器是從@Pixel 的答案中復制的。 我只是把它放在一起。

public class DecimalDigitsInputFilter implements InputFilter {

    Pattern mPattern;

    public DecimalDigitsInputFilter() {
        mPattern = Pattern.compile("([1-9]{1}[0-9]{0,2}([0-9]{3})*(\\.[0-9]{0,2})?|[1-9]{1}[0-9]{0,}(\\.[0-9]{0,2})?|0(\\.[0-9]{0,2})?|(\\.[0-9]{1,2})?)");

    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {

        String formatedSource = source.subSequence(start, end).toString();

        String destPrefix = dest.subSequence(0, dstart).toString();

        String destSuffix = dest.subSequence(dend, dest.length()).toString();

        String result = destPrefix + formatedSource + destSuffix;

        result = result.replace(",", ".");

        Matcher matcher = mPattern.matcher(result);

        if (matcher.matches()) {
            return null;
        }

        return "";
    }
}

現在像這樣在EditText設置過濾器。

mEditText.setFilters(new InputFilter[]{new DecimalDigitsInputFilter()});

這里一件重要的事情是它確實解決了我的問題，即不允許在該EditText中顯示小數點后兩位以上的數字，但問題是當我從EditText getText() ，它返回我輸入的整個輸入。

例如，在EditText應用過濾器后，我嘗試設置輸入 1.5699856987。 所以在屏幕上它顯示了 1.56，這是完美的。

然后我想將此輸入用於其他一些計算，因此我想從該輸入字段 ( EditText ) 中獲取文本。 當我調用mEditText.getText().toString()它返回 1.5699856987 這在我的情況下是不可接受的。

所以我必須在從EditText獲取它后再次解析該值。

BigDecimal amount = new BigDecimal(Double.parseDouble(mEditText.getText().toString().trim()))
    .setScale(2, RoundingMode.HALF_UP);

在從EditText獲取全文后， setScale在此處執行此操作。

Answer 22

在 Android kotlin 中創建一個名為 DecimalDigitsInputFilter 的新類

class DecimalDigitsInputFilter(digitsBeforeDecimal: Int, digitsAfterDecimal: Int) : InputFilter {

    var mPattern: Pattern = Pattern.compile("[0-9]{0,$digitsBeforeDecimal}+((\\.[0-9]{0,$digitsAfterDecimal})?)||(\\.)?")

    override fun filter(
        source: CharSequence?,
        start: Int,
        end: Int,
        dest: Spanned?,
        dstart: Int,
        dend: Int
    ): CharSequence? {
        val matcher: Matcher = mPattern.matcher(
            dest?.subSequence(0, dstart).toString() + source?.subSequence(
                start,
                end
            ).toString() + dest?.subSequence(dend, dest.length).toString()
        )
        if (!matcher.matches())
            return ""
        else
            return null
    }
}

使用以下行調用此類

 et_buy_amount.filters = (arrayOf<InputFilter>(DecimalDigitsInputFilter(8,2)))

有太多相同的答案，但它允許您輸入小數點前 8 位和小數點后 2 位

其他答案只接受 8 位數字

Answer 23

我也遇到過這個問題。 我希望能夠在許多 EditText 中重用代碼。 這是我的解決方案：

用法 ：

CurrencyFormat watcher = new CurrencyFormat();
priceEditText.addTextChangedListener(watcher);

班級：

public static class CurrencyFormat implements TextWatcher {

    public void onTextChanged(CharSequence arg0, int start, int arg2,int arg3) {}

    public void beforeTextChanged(CharSequence arg0, int start,int arg2, int arg3) {}

    public void afterTextChanged(Editable arg0) {
        int length = arg0.length();
        if(length>0){
            if(nrOfDecimal(arg0.toString())>2)
                    arg0.delete(length-1, length);
        }

    }


    private int nrOfDecimal(String nr){
        int len = nr.length();
        int pos = len;
        for(int i=0 ; i<len; i++){
            if(nr.charAt(i)=='.'){
                pos=i+1;
                    break;
            }
        }
        return len-pos;
    }
}

Answer 24

@Meh 給你..

txtlist.setFilters(new InputFilter[] { new DigitsKeyListener( Boolean.FALSE,Boolean.TRUE) {

        int beforeDecimal = 7;
        int afterDecimal = 2;

        @Override
        public CharSequence filter(CharSequence source, int start, int end,Spanned dest, int dstart, int dend) {

            String etText = txtlist.getText().toString();
            String temp = txtlist.getText() + source.toString();
            if (temp.equals(".")) {
                return "0.";
            } else if (temp.toString().indexOf(".") == -1) {
                // no decimal point placed yet
                 if (temp.length() > beforeDecimal) {
                    return "";
                }
            } else {
                int dotPosition ;
                int cursorPositon = txtlistprice.getSelectionStart();
                if (etText.indexOf(".") == -1) {
                    dotPosition = temp.indexOf(".");
                }else{
                    dotPosition = etText.indexOf(".");
                }
                if(cursorPositon <= dotPosition){
                    String beforeDot = etText.substring(0, dotPosition);
                    if(beforeDot.length()<beforeDecimal){
                        return source;
                    }else{
                        if(source.toString().equalsIgnoreCase(".")){
                            return source;
                        }else{
                            return "";
                        }
                    }
                }else{
                    temp = temp.substring(temp.indexOf(".") + 1);
                    if (temp.length() > afterDecimal) {
                        return "";
                    }
                }
            }
            return super.filter(source, start, end, dest, dstart, dend);
        }
    } });

Answer 25

這是TextWatcher ， TextWatcher允許小數點后有n個數字。

文本觀察者

private static boolean flag;
public static TextWatcher getTextWatcherAllowAfterDeci(final int allowAfterDecimal){

    TextWatcher watcher = new TextWatcher() {

        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {
            // TODO Auto-generated method stub

        }

        @Override
        public void beforeTextChanged(CharSequence s, int start, int count,
                int after) {
            // TODO Auto-generated method stub

        }

        @Override
        public void afterTextChanged(Editable s) {
            // TODO Auto-generated method stub
            String str = s.toString();
            int index = str.indexOf ( "." );
            if(index>=0){
                if((index+1)<str.length()){
                    String numberD = str.substring(index+1);
                    if (numberD.length()!=allowAfterDecimal) {
                        flag=true;
                    }else{
                        flag=false;
                    }   
                }else{
                    flag = false;
                }                   
            }else{
                flag=false;
            }
            if(flag)
                s.delete(s.length() - 1,
                        s.length());
        }
    };
    return watcher;
}

如何使用

yourEditText.addTextChangedListener(getTextWatcherAllowAfterDeci(1));

Answer 26

一個很晚的回應：我們可以簡單地這樣做：

etv.addTextChangedListener(new TextWatcher() {
        @Override
        public void beforeTextChanged(CharSequence s, int start, int count, int after) {
        }

        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {
            if (s.toString().length() > 3 && s.toString().contains(".")) {
                if (s.toString().length() - s.toString().indexOf(".") > 3) {
                    etv.setText(s.toString().substring(0, s.length() - 1));
                    etv.setSelection(edtSendMoney.getText().length());
                }
            }
        }

        @Override
        public void afterTextChanged(Editable arg0) {
        }
}

Answer 27

這是我的解決方案：

     yourEditText.addTextChangedListener(new TextWatcher() {
        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {
            NumberFormat formatter = new DecimalFormat("#.##");
            double doubleVal = Double.parseDouble(s.toString());
            yourEditText.setText(formatter.format(doubleVal));
        }

        @Override
        public void beforeTextChanged(CharSequence s, int start, int count,int after) {}

        @Override
        public void afterTextChanged(Editable s) {}
    });

如果用戶輸入的數字小數點后超過兩個數字，則會自動更正。

我希望我有幫助！

Answer 28

et = (EditText) vw.findViewById(R.id.tx_edittext);

et.setFilters(new InputFilter[] {
        new DigitsKeyListener(Boolean.FALSE, Boolean.TRUE) {
            int beforeDecimal = 5, afterDecimal = 2;

            @Override
            public CharSequence filter(CharSequence source, int start, int end,
                    Spanned dest, int dstart, int dend) {
                String temp = et.getText() + source.toString();

                if (temp.equals(".")) {
                    return "0.";
                }
                else if (temp.toString().indexOf(".") == -1) {
                    // no decimal point placed yet
                    if (temp.length() > beforeDecimal) {
                        return "";
                    }
                } else {
                    temp = temp.substring(temp.indexOf(".") + 1);
                    if (temp.length() > afterDecimal) {
                        return "";
                    }
                }

                return super.filter(source, start, end, dest, dstart, dend);
            }
        }
});

Answer 29

這對我來說很好用。 它允許即使在焦點更改並檢索回來后也可以輸入值。 例如： 123.00 、 12.12 、 0.01等。

1. Integer.parseInt(getString(R.string.valuelength))指定輸入digits.Values的長度。從string.xml文件訪問的值。改變值很安靜容易。 2. Integer.parseInt(getString(R.string.valuedecimal)) ，這是小數位數的最大限制。

private InputFilter[] valDecimalPlaces;
private ArrayList<EditText> edittextArray;

valDecimalPlaces = new InputFilter[] { new DecimalDigitsInputFilterNew(
    Integer.parseInt(getString(R.string.valuelength)),
    Integer.parseInt(getString(R.string.valuedecimal))) 
};

允許執行操作的EditText值數組。

for (EditText etDecimalPlace : edittextArray) {
            etDecimalPlace.setFilters(valDecimalPlaces);

我只是使用了包含多個 edittext Next DecimalDigitsInputFilterNew.class文件的值數組。

import android.text.InputFilter;
import android.text.Spanned;

public class DecimalDigitsInputFilterNew implements InputFilter {

    private final int decimalDigits;
    private final int before;

    public DecimalDigitsInputFilterNew(int before ,int decimalDigits) {
        this.decimalDigits = decimalDigits;
        this.before = before;
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end,
        Spanned dest, int dstart, int dend) {
        StringBuilder builder = new StringBuilder(dest);
        builder.replace(dstart, dend, source
              .subSequence(start, end).toString());
        if (!builder.toString().matches("(([0-9]{1})([0-9]{0,"+(before-1)+"})?)?(\\.[0-9]{0,"+decimalDigits+"})?")) {
             if(source.length()==0)
                  return dest.subSequence(dstart, dend);
             return "";
        }
        return null;
    }
}

Answer 30

這是建立在 pinhassi 的回答之上的 - 我遇到的問題是，一旦達到小數限制，您就無法在小數點之前添加值。 為了解決這個問題，我們需要在進行模式匹配之前構造最終的字符串。

import java.util.regex.Matcher;
import java.util.regex.Pattern;

import android.text.InputFilter;
import android.text.Spanned;

public class DecimalLimiter implements InputFilter
{
    Pattern mPattern;

    public DecimalLimiter(int digitsBeforeZero,int digitsAfterZero) 
    {
        mPattern=Pattern.compile("[0-9]{0," + (digitsBeforeZero) + "}+((\\.[0-9]{0," + (digitsAfterZero) + "})?)||(\\.)?");
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) 
    {
        StringBuilder sb = new StringBuilder(dest);
        sb.insert(dstart, source, start, end);

        Matcher matcher = mPattern.matcher(sb.toString());
        if(!matcher.matches())
            return "";
        return null;
    }
}

Answer 31

這段代碼很好用，

public class DecimalDigitsInputFilter implements InputFilter {

    private final int digitsBeforeZero;
    private final int digitsAfterZero;
    private Pattern mPattern;

    public DecimalDigitsInputFilter(int digitsBeforeZero, int digitsAfterZero) {
        this.digitsBeforeZero = digitsBeforeZero;
        this.digitsAfterZero = digitsAfterZero;
        applyPattern(digitsBeforeZero, digitsAfterZero);
    }

    private void applyPattern(int digitsBeforeZero, int digitsAfterZero) {
        mPattern = Pattern.compile("[0-9]{0," + (digitsBeforeZero - 1) + "}+((\\.[0-9]{0," + (digitsAfterZero - 1) + "})?)|(\\.)?");
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
        if (dest.toString().contains(".") || source.toString().contains("."))
            applyPattern(digitsBeforeZero + 2, digitsAfterZero);
        else
            applyPattern(digitsBeforeZero, digitsAfterZero);

        Matcher matcher = mPattern.matcher(dest);
        if (!matcher.matches())
            return "";
        return null;
    }

}

應用過濾器：

edittext.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});

Answer 32

就像其他人說的，我在我的項目中添加了這個類並將過濾器設置為 EditText Simpler 解決方案，而不使用正則表達式：

public class DecimalDigitsInputFilter implements InputFilter {
int digitsBeforeZero =0;
int digitsAfterZero=0;

public DecimalDigitsInputFilter(int digitsBeforeZero,int digitsAfterZero) {
this.digitsBeforeZero=digitsBeforeZero;
this.digitsAfterZero=digitsAfterZero;
}

@Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
if(dest!=null && dest.toString().trim().length()<(digitsBeforeZero+digitsAfterZero)){
    String value=dest.toString().trim();
    if(value.contains(".") && (value.substring(value.indexOf(".")).length()<(digitsAfterZero+1))){
        return ((value.indexOf(".")+1+digitsAfterZero)>dstart)?null:"";
    }else if(value.contains(".") && (value.indexOf(".")<dstart)){
        return "";
    }else if(source!=null && source.equals(".")&& ((value.length()-dstart)>=(digitsAfterZero+1))){
        return "";
    }

}else{
    return "";
}
    return null;
}

}

應用過濾器：

edittext.setFilters(new InputFilter[] {new DecimalDigitsInputFilter(5,2)});

Answer 33

我沒有正則表達式的簡單解決方案

    int start=Edit1.getSelectionStart();
    String sp=Edit1.getText().toString();
    sp=sp.replace(",",".");
    Double d=Double.valueOf(sp);
    String s=String.format("%.2f",d );
    if(!Edit1.getText().toString().equals(s))
        Edit1.setText(s);
    if(start>Edit1.getText().length())start--;
    Edit1.setSelection(start);

放入 onTextChange。 通過刪除逗號，零會加倍，因為數字變成了整數。

Answer 34

@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
    String numero = total.getText().toString();
    int dec = numero.indexOf(".");
    int longitud = numero.length();
    if (dec+3 == longitud && dec != -1) { //3 number decimal + 1
        log.i("ento","si");
        numero = numero.substring(0,dec+3);
        if (contador == 0) {
            contador = 1;
            total.setText(numero);
            total.setSelection(numero.length());
        } else {
            contador = 0;
        }
    }
}

Answer 35

InputFilter 的這個實現解決了這個問題。

import android.text.SpannableStringBuilder;
import android.text.Spanned;
import android.text.method.DigitsKeyListener;

public class MoneyValueFilter extends DigitsKeyListener {
    public MoneyValueFilter() {
        super(false, true);
    }

    private int digits = 2;

    public void setDigits(int d) {
        digits = d;
    }

    @Override
    public CharSequence filter(CharSequence source, int start, int end,
            Spanned dest, int dstart, int dend) {
        CharSequence out = super.filter(source, start, end, dest, dstart, dend);

        // if changed, replace the source
        if (out != null) {
            source = out;
            start = 0;
            end = out.length();
        }

        int len = end - start;

        // if deleting, source is empty
        // and deleting can't break anything
        if (len == 0) {
            return source;
        }

        int dlen = dest.length();

        // Find the position of the decimal .
        for (int i = 0; i < dstart; i++) {
            if (dest.charAt(i) == '.') {
                // being here means, that a number has
                // been inserted after the dot
                // check if the amount of digits is right
                return (dlen-(i+1) + len > digits) ? 
                    "" :
                    new SpannableStringBuilder(source, start, end);
            }
        }

        for (int i = start; i < end; ++i) {
            if (source.charAt(i) == '.') {
                // being here means, dot has been inserted
                // check if the amount of digits is right
                if ((dlen-dend) + (end-(i + 1)) > digits)
                    return "";
                else
                    break;  // return new SpannableStringBuilder(source, start, end);
            }
        }

        // if the dot is after the inserted part,
        // nothing can break
        return new SpannableStringBuilder(source, start, end);
    }
}

使用：

editCoin.setFilters(new InputFilter[] {new MoneyValueFilter(2)});

Answer 36

Kotlin 中的簡單 BindingAdapter：

@BindingAdapter("maxDecimalPlaces")
fun TextInputEditText.limitDecimalPlaces(maxDecimalPlaces: Int) {
    filters += InputFilter { source, _, _, dest, dstart, dend ->
        val value = if (source.isEmpty()) {
            dest.removeRange(dstart, dend)
        } else {
            StringBuilder(dest).insert(dstart, source)
        }
        val matcher = Pattern.compile("([1-9][0-9]*)|([1-9][0-9]*\\.[0-9]{0,$maxDecimalPlaces})|(\\.[0-9]{0,$maxDecimalPlaces})").matcher(value)
        if (!matcher.matches()) "" else null
    }
}

Answer 37

如果你想對整數部分有限制，這里也是代碼

class PropertyCostInputFilter : DigitsKeyListener(false, true) {

override fun filter(source: CharSequence, start: Int, end: Int, dest: Spanned, dstart: Int, dend: Int): CharSequence {
    var source = source
    var start = start
    var end = end
    val out = super.filter(source, start, end, dest, dstart, dend)

    if (out != null) {
        source = out
        start = 0
        end = out.length
    }

    val sourceLength = end - start

    // If length = 0, then there was a deletion and therefore the length could not become greater than the max value
    if (sourceLength == 0) {
        return source
    }

    val result = dest.replaceRange((dstart until dend), source.substring(start, end))
    val parts = result.split(SEPARATOR)

    if (parts.size > 0 && parts[0].length > INTEGER_PART_MAX_DIGITS
        || parts.size > 1 && parts[1].length > FRACTIONAL_PART_MAX_DIGITS
    ) {
        return ""
    }

    return SpannableStringBuilder(source, start, end)
}

companion object {
    private const val INTEGER_PART_MAX_DIGITS = 20
    private const val FRACTIONAL_PART_MAX_DIGITS = 2
    private const val SEPARATOR = '.'
}
}

Answer 38

對於科特林

val inputFilter =  arrayOf<InputFilter>(DecimalDigitsInputFilter(5,2))
            et_total_value.setFilters(inputFilter)

Answer 39

這是將小數點后的位數限制為兩位的最簡單解決方案：

myeditText2 = (EditText) findViewById(R.id.editText2);  
myeditText2.setInputType(3);