從 android 的編輯文本框中獲取 XXX-XXX-XXXX 格式的電話號碼

Question

嗨，我正在寫一個 android，其中用戶在編輯文本框中輸入電話號碼。 我希望數字采用 xxx-xxx-xxxx 的形式，這意味着在用戶輸入前 3 個字母和另一個 '-' 后，'-' 應該自動出現

我用EditText anum= (EditText)findViewById(R.id.altnum); anum.addTextChangedListener(new PhoneNumberFormattingTextWatcher()); EditText anum= (EditText)findViewById(R.id.altnum); anum.addTextChangedListener(new PhoneNumberFormattingTextWatcher());

但只有在輸入所有數字后才會進入格式。 我希望在用戶輸入數據時進行更改，例如如果他按 123，hypen 應該會自動出現請告訴我該怎么做。

感謝您

此致

ChinniKrishna Kothapalli

Answer 1

最近，我也有同樣的要求。 我用 TextWatcher 試過這個。 在這里分享，希望其他人以后可能需要。

public class PhoneNumberTextWatcher implements TextWatcher {

private static final String TAG = PhoneNumberTextWatcher.class
        .getSimpleName();
private EditText edTxt;
private boolean isDelete;

public PhoneNumberTextWatcher(EditText edTxtPhone) {
    this.edTxt = edTxtPhone;
    edTxt.setOnKeyListener(new OnKeyListener() {

        @Override
        public boolean onKey(View v, int keyCode, KeyEvent event) {
            if (keyCode == KeyEvent.KEYCODE_DEL) {
                isDelete = true;
            }
            return false;
        }
    });
}

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

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

public void afterTextChanged(Editable s) {

    if (isDelete) {
        isDelete = false;
        return;
    }
    String val = s.toString();
    String a = "";
    String b = "";
    String c = "";
    if (val != null && val.length() > 0) {
        val = val.replace("-", "");
        if (val.length() >= 3) {
            a = val.substring(0, 3);
        } else if (val.length() < 3) {
            a = val.substring(0, val.length());
        }
        if (val.length() >= 6) {
            b = val.substring(3, 6);
            c = val.substring(6, val.length());
        } else if (val.length() > 3 && val.length() < 6) {
            b = val.substring(3, val.length());
        }
        StringBuffer stringBuffer = new StringBuffer();
        if (a != null && a.length() > 0) {
            stringBuffer.append(a);
            if (a.length() == 3) {
                stringBuffer.append("-");
            }
        }
        if (b != null && b.length() > 0) {
            stringBuffer.append(b);
            if (b.length() == 3) {
                stringBuffer.append("-");
            }
        }
        if (c != null && c.length() > 0) {
            stringBuffer.append(c);
        }
        edTxt.removeTextChangedListener(this);
        edTxt.setText(stringBuffer.toString());
        edTxt.setSelection(edTxt.getText().toString().length());
        edTxt.addTextChangedListener(this);
    } else {
        edTxt.removeTextChangedListener(this);
        edTxt.setText("");
        edTxt.addTextChangedListener(this);
    }

}
 }

謝謝

JRH

Answer 2

以下代碼可以很好地添加和刪除場景，但邏輯要長一些：

public class PhoneNumberTextWatcher implements TextWatcher {

private static final String TAG = "PhoneNumberTextWatcher";
private EditText editText;

public PhoneNumberTextWatcher(EditText edTxtPhone) {
    this.editText = edTxtPhone;
}

public void onTextChanged(CharSequence s, int cursorPosition, int before,
                          int count) {

    if(before == 0 && count == 1){  //Entering values

        String val = s.toString();
        String a = "";
        String b = "";
        String c = "";
        if (val != null && val.length() > 0) {
            val = val.replace("-", "");
            if (val.length() >= 3) {
                a = val.substring(0, 3);
            } else if (val.length() < 3) {
                a = val.substring(0, val.length());
            }
            if (val.length() >= 6) {
                b = val.substring(3, 6);
                c = val.substring(6, val.length());
            } else if (val.length() > 3 && val.length() < 6) {
                b = val.substring(3, val.length());
            }
            StringBuffer stringBuffer = new StringBuffer();
            if (a != null && a.length() > 0) {
                stringBuffer.append(a);

            }
            if (b != null && b.length() > 0) {
                stringBuffer.append("-");
                stringBuffer.append(b);

            }
            if (c != null && c.length() > 0) {
                stringBuffer.append("-");
                stringBuffer.append(c);
            }
            editText.removeTextChangedListener(this);
            editText.setText(stringBuffer.toString());
            if(cursorPosition == 3 || cursorPosition == 7){
                cursorPosition = cursorPosition+2;
            }else{
                cursorPosition = cursorPosition+1;
            }
            if(cursorPosition <= editText.getText().toString().length()) {
                editText.setSelection(cursorPosition);
            }else{
                editText.setSelection(editText.getText().toString().length());
            }
            editText.addTextChangedListener(this);
        } else {
            editText.removeTextChangedListener(this);
            editText.setText("");
            editText.addTextChangedListener(this);
        }

    }

    if(before == 1 && count == 0){  //Deleting values

        String val = s.toString();
        String a = "";
        String b = "";
        String c = "";

        if (val != null && val.length() > 0) {
            val = val.replace("-", "");
            if(cursorPosition == 3){
                val = removeCharAt(val,cursorPosition-1,s.toString().length()-1);
            }else if(cursorPosition == 7){
                val = removeCharAt(val,cursorPosition-2,s.toString().length()-2);
            }
            if (val.length() >= 3) {
                a = val.substring(0, 3);
            } else if (val.length() < 3) {
                a = val.substring(0, val.length());
            }
            if (val.length() >= 6) {
                b = val.substring(3, 6);
                c = val.substring(6, val.length());
            } else if (val.length() > 3 && val.length() < 6) {
                b = val.substring(3, val.length());
            }
            StringBuffer stringBuffer = new StringBuffer();
            if (a != null && a.length() > 0) {
                stringBuffer.append(a);

            }
            if (b != null && b.length() > 0) {
                stringBuffer.append("-");
                stringBuffer.append(b);

            }
            if (c != null && c.length() > 0) {
                stringBuffer.append("-");
                stringBuffer.append(c);
            }
            editText.removeTextChangedListener(this);
            editText.setText(stringBuffer.toString());
            if(cursorPosition == 3 || cursorPosition == 7){
                cursorPosition = cursorPosition-1;
            }
            if(cursorPosition <= editText.getText().toString().length()) {
                editText.setSelection(cursorPosition);
            }else{
                editText.setSelection(editText.getText().toString().length());
            }
            editText.addTextChangedListener(this);
        } else {
            editText.removeTextChangedListener(this);
            editText.setText("");
            editText.addTextChangedListener(this);
        }

    }



}

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

public void afterTextChanged(Editable s) {


}

public static String removeCharAt(String s, int pos,int length) {

    String value = "";
    if(length > pos){
        value = s.substring(pos + 1);
    }
    return s.substring(0, pos)+value ;
}
}

Answer 3

一種選擇是實現您自己的InputFilter 。

你可以在這里使用我的答案： 按“。” 多次（在鍵入時在 EditText 中驗證 ip 地址）和此處： 如何設置 Edittext 視圖僅允許兩個數字值和兩個十進制值，例如 ##.## ，例如如何在鍵入時解析文本。

如果您希望破折號自動出現，您需要將它們添加到您的過濾器方法的返回