How to remove only trailing spaces of a string in Java and keep leading spaces?

Solution 1:

Since JDK 11

If you are on JDK 11 or higher you should probably be using stripTrailing().

Earlier JDK versions

Using the regular expression \s++$, you can replace all trailing space characters (includes space and tab characters) with the empty string ("").

final String text = "  foo   ";
System.out.println(text.replaceFirst("\\s++$", ""));

Output

foo

Online demo.

Here's a breakdown of the regex:

\s – any whitespace character,
++ – match one or more of the previous token (possessively); i.e., match one or more whitespace character. The + pattern is used in its possessive form ++, which takes less time to detect the case when the pattern does not match.
$ – the end of the string.

Thus, the regular expression will match as much whitespace as it can that is followed directly by the end of the string: in other words, the trailing whitespace.

The investment into learning regular expressions will become more valuable, if you need to extend your requirements later on.

References

Java regular expression syntax

Solution 2:

Another option is to use Apache Commons StringUtils, specifically StringUtils.stripEnd

String stripped = StringUtils.stripEnd("   my lousy string    "," ");

Solution 3:

I modified the original java.lang.String.trim() method a bit and it should work:

  public String trim(String str) {
        int len = str.length();
        int st = 0;

        char[] val = str.toCharArray();

        while ((st < len) && (val[len - 1] <= ' ')) {
            len--;
        }
        return str.substring(st, len);
    }

Test:

  Test test = new Test();
  String sample = "            Hello World               "; // A String with trailing and leading spaces
  System.out.println(test.trim(sample) + " // No trailing spaces left");

Output:

        Hello World // No trailing spaces left