Regular expression to replace first 4 characters

Question

I am trying to write a Regular Expression that replaces the first 4 characters of a string with *s

For example, for the 123456 input, the expected output is ****56.

If the input length is less than 4 then return only *s.

For example, if the input is 123, the return value must be ***.

Answer 1

Here is a simple solution using String#repeat(int) available as of java-11. Regex is not needed.

static String hideFirst(String string, int size) {
    return size < string.length() ?                     // if string is longer than size
            "*".repeat(size) + string.substring(size) : // ... replace the part
            "*".repeat(string.length());                // ... or replace the whole
}

String s1 = hideFirst("123456789", 4);    // ****56789
String s2 = hideFirst("12345", 4);        // ****5
String s3 = hideFirst("1234", 4);         // ****
String s4 = hideFirst("123", 4);          // ***
String s5 = hideFirst("", 4);             // (empty string)

• You might want to pass the asterisk (or any) character to the method for more control
• You might want to handle null (return null / throw the NPE / throw a custom exception...)
• For lower versions of Java, you need a different approach for the String repetition.