在android中的textview中过滤文本

Question

Is there a way to remove integers from TextView in android. For example let's say we have text like this:

123Isuru456Ranasinghe

I want this text to be like this after removing integers

IsuruRanasinghe

How do I achieve this in android?

Answer 1

This will Help you.

  public static String removeDigits(String text) {
        int length = text.length();
        StringBuffer buffer = new StringBuffer(length);
        for(int i = 0; i < length; i++) {
            char ch = text.charAt(i);
            if (!Character.isDigit(ch)) {
                buffer.append(ch);
            }
        }
        return buffer.toString();
    }

Another Simple Option :

// do it in just one line of code
String num = text.replaceAll(”[\\d]“, “”);

Return your string with Removing Digits.

Answer 2

This is just pure java. Nothing to do with Android.
Here is the code to do what you want.

String str = "123Isuru456Ranasinghe";
String newStr = str.replaceAll("[0-9]", "");

After some tests, it seems that the longest solution is the best in the matter of performance !

public static void main(String[] arg) throws IOException {
    // Building a long string...
    StringBuilder str = new StringBuilder();
    for (int i = 0; i < 1000000; i++)
        str.append("123Isuru456Ranasinghe");

    removeNum1(str.toString());
    removeNum2(str.toString());
}

// With a replaceAll => 1743 ms
private static void removeNum1(String _str) {
    long start = System.currentTimeMillis();
    String finalString = _str.replaceAll("[0-9]", "");
    System.out.println(System.currentTimeMillis() - start);
}

// With StringBuilder and loop => 348 ms
private static void removeNum2(String _str) {
    long start = System.currentTimeMillis();

    StringBuilder finalString = new StringBuilder();
    char currentChar;
    for (int i = 0; i < _str.length(); ++i) {
        currentChar = _str.charAt(i);
        if (Character.isLetter(currentChar)) {
            finalString.append(currentChar);
        }
    }
    System.out.println(System.currentTimeMillis() - start);
}

Using a loop is much faster. But in your case, it is a bit useless :p

Now you have to choose between "slow" and short to write and very fast but a bit more complicated. All depend of what you need.

Answer 3

StringBuilder ans = new StringBuilder();
char currentChar;
for (int i = 0; i < str.length(); ++i) {
    currentChar = str.charAt(i);
    if (Character.isLetter(currentChar)) {
        ans.append(currentChar);
    }
}