'How to make Textview will only show the last 4 digits numbers

My Textview is getting the data from SQLite, but I want to make Textview to show as follows:
************1234, instead of 1234123412341234.

enter image description here

javaandroidxml


Solution 1:[1]

public static String padLeading*(String data, int requiredLength){
        StringBuffer sb = new StringBuffer();

        if(data .length() > 4){
data .substring(data .length() - 4)
}

        int numLeading* = requiredLength - data.length();
        for (int i=0; i < numLeading*; i++){
            sb.append("*");
        }
        sb.append(data);
        return sb.toString();
    }

Advantage of this method is that you could vary the number of * as required. Or to make it simpler could determine the last four characters of the string, pass it in and take:

if(data .length() > 4){
    data .substring(data .length() - 4)
    }

out of this method.

Solution 2:[2]

Kotlin solution

acNumber.replace("\\w(?=\\w{4})".toRegex(),"*")

Solution 3:[3]

String string = "1234567890";
if(string .length() > 4){
string .substring(string .length() - 4)
}
tv.setText(string);

output: 7890

Solution 4:[4]

You can hard code "*" in the beginning and extract last 4 digits of credit card num from sqlite and display on text view.

String lastFourDigits = extractLastFourDigits(); //I assume you get this
myTextView.setText("**** **** **** "+lastFourDigits);

Solution 5:[5]

You want to use a TransformationMethod You can use the source of PasswordTransformationMethod as a guide so you don't lose the original string.

Solution 6:[6]

Here you go. Clean and reusable:

/**
 * Applies the specified mask to the card number.
 *
 * @param cardNumber The card number in plain format
 * @param mask The number mask pattern. Use # to include a digit from the
 * card number at that position, use x to skip the digit at that position
 *
 * @return The masked card number
 */
public static String maskCardNumber(String cardNumber, String mask) {

    // format the number
    int index = 0;
    StringBuilder maskedNumber = new StringBuilder();
    for (int i = 0; i < mask.length(); i++) {
        char c = mask.charAt(i);
        if (c == '#') {
            maskedNumber.append(cardNumber.charAt(index));
            index++;
        } else if (c == 'x') {
            maskedNumber.append(c);
            index++;
        } else {
            maskedNumber.append(c);
        }
    }

    // return the masked number
    return maskedNumber.toString();
}

Sample Calls:

System.out.println(maskCardNumber("1234123412341234", "xxxx-xxxx-xxxx-####"));
> xxxx-xxxx-xxxx-1234

System.out.println(maskCardNumber("1234123412341234", "##xx-xxxx-xxxx-xx##"));
> 12xx-xxxx-xxxx-xx34

Solution 7:[7]

stringData.replaceRange(0, stringData.length - 2, "*")

I hope it will work

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 user8537453
Solution 2 Vikash Parajuli
Solution 3 Yoni
Solution 4 Aalap Patel
Solution 5 Ge3ng
Solution 6 AbhinayMe
Solution 7

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