Remove end of line characters from end of Java String

Question

I have a string which I'd like to remove the end of line characters from the very end of the string only using Java

"foo\r\nbar\r\nhello\r\nworld\r\n"

which I'd like to become

"foo\r\nbar\r\nhello\r\nworld"

(This question is similar to, but not the same as question 593671)

Answer 1

You can use s = s.replaceAll("[\r\n]+$", "");. This trims the \r and \n characters at the end of the string

The regex is explained as follows:

• [\r\n] is a character class containing \r and \n
• + is one-or-more repetition of
• $ is the end-of-string anchor

References

• regular-expressions.info/Anchors, Character Class, Repetition

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Related topics

You can also use String.trim() to trim any whitespace characters from the beginning and end of the string:

s = s.trim();

If you need to check if a String contains nothing but whitespace characters, you can check if it isEmpty() after trim():

if (s.trim().isEmpty()) {
   //...
}

Alternatively you can also see if it matches("\\s*"), i.e. zero-or-more of whitespace characters. Note that in Java, the regex matches tries to match the whole string. In flavors that can match a substring, you need to anchor the pattern, so it's ^\s*$.

Related questions

• regex, check if a line is blank or not
• how to replace 2 or more spaces with single space in string and delete leading spaces only

Answer 2

Wouldn't String.trim do the trick here?

i.e you'd call the method .trim() on your string and it should return a copy of that string minus any leading or trailing whitespace.

Answer 3

The Apache Commons Lang StringUtils.stripEnd(String str, String stripChars) will do the trick; e.g.

    String trimmed = StringUtils.stripEnd(someString, "\n\r");

If you want to remove all whitespace at the end of the String:

    String trimmed = StringUtils.stripEnd(someString, null);

Answer 4

Well, everyone gave some way to do it with regex, so I'll give a fastest way possible instead:

public String replace(String val) {
    for (int i=val.length()-1;i>=0;i--) {
        char c = val.charAt(i);
        if (c != '\n' && c != '\r') {
            return val.substring(0, i+1);
        }
    }
    return "";
}

Benchmark says it operates ~45 times faster than regexp solutions.

Answer 5

If you have Google's guava-librariesin your project (if not, you arguably should!) you'd do this with a CharMatcher:

String result = CharMatcher.any("\r\n").trimTrailingFrom(input);

Answer 6

String text = "foo\r\nbar\r\nhello\r\nworld\r\n";
String result = text.replaceAll("[\r\n]+$", "");

Answer 7

"foo\r\nbar\r\nhello\r\nworld\r\n".replaceAll("\\s+$", "")
or
"foo\r\nbar\r\nhello\r\nworld\r\n".replaceAll("[\r\n]+$", "")