I'm writing a binary/decimal/hexadecimal converter for Android and am trying to format a String in Java with a regex that will add a space to every four characters from right to left.
This code works from left to right, but I am wondering if there is a way I can reverse it.
stringNum = stringNum.replaceAll("....", "$0 ");
Maybe instead of regex use StringBuilder like this
String data = "abcdefghij";
StringBuilder sb = new StringBuilder(data);
for (int i = sb.length() - 4; i > 0; i -= 4)
sb.insert(i, ' ');
data = sb.toString();
System.out.println(data);
output:
ab cdef ghij
In C# you would do something like Regex.Replace("1234567890", ".{4}", " $0", RegexOptions.RightToLeft).
Unfortunately, in Java you can't.
As Explossion Pills suggested in a comment, you could reverse the string, apply the regex, and reverse it again.
Another solution could be to take the last N numbers from your string, where N is divisible by 4. Then you can apply the replacement successfully, and concatenate the remaining part of the string at the begining.
String str = "0123456789ABCDE";
// Use a StringBuffer to reverse the string
StringBuffer sb = new StringBuffer(str).reverse();
// This regex causes " " to occur before each instance of four characters
str = sb.toString().replaceAll("....", "$0 ");
// Reverse it again
str = (new StringBuffer(str)).reverse().toString();
System.out.println(str);
For me the best practice was to start from the beginning of the string to the end:
String inputString = "1234123412341234";
int j = 0;
StringBuilder sb = new StringBuilder(inputString);
for (int i = 4; i < kkm.getFiscalNumber().length(); i += 4) {
sb.insert(i + j, ' ');
j++;
}
String result = sb.toString(); // 1234 1234 1234 1234
