How to remove newlines from beginning and end of a string?

I have a string that contains some text followed by a blank line. What's the best way to keep the part with text, but remove the whitespace newline from the end?

Use String.trim() method to get rid of whitespaces (spaces, new lines etc.) from the beginning and end of the string.

String trimmedString = myString.trim();

String.replaceAll("[\n\r]", "");

This Java code does exactly what is asked in the title of the question, that is "remove newlines from beginning and end of a string-java":

String.replaceAll("^[\n\r]", "").replaceAll("[\n\r]$", "")

Remove newlines only from the end of the line:

String.replaceAll("[\n\r]$", "")

Remove newlines only from the beginning of the line:

String.replaceAll("^[\n\r]", "")

tl;dr

String cleanString = dirtyString.strip() ; // Call new `String::string` method.

String::strip…

The old String::trim method has a strange definition of whitespace.

As discussed here, Java 11 adds new strip… methods to the String class. These use a more Unicode-savvy definition of whitespace. See the rules of this definition in the class JavaDoc for Character::isWhitespace.

Example code.

String input = " some Thing ";
System.out.println("before->>"+input+"<<-");
input = input.strip();
System.out.println("after->>"+input+"<<-");

Or you can strip just the leading or just the trailing whitespace.

You do not mention exactly what code point(s) make up your newlines. I imagine your newline is likely included in this list of code points targeted by strip:

• It is a Unicode space character (SPACE_SEPARATOR, LINE_SEPARATOR, or PARAGRAPH_SEPARATOR) but is not also a non-breaking space ('\u00A0', '\u2007', '\u202F').
• It is '\t', U+0009 HORIZONTAL TABULATION.
• It is '\n', U+000A LINE FEED.
• It is '\u000B', U+000B VERTICAL TABULATION.
• It is '\f', U+000C FORM FEED.
• It is '\r', U+000D CARRIAGE RETURN.
• It is '\u001C', U+001C FILE SEPARATOR.
• It is '\u001D', U+001D GROUP SEPARATOR.
• It is '\u001E', U+001E RECORD SEPARATOR.
• It is '\u001F', U+0

If your string is potentially null, consider using StringUtils.trim() - the null-safe version of String.trim().

If you only want to remove line breaks (not spaces, tabs) at the beginning and end of a String (not inbetween), then you can use this approach:

Use a regular expressions to remove carriage returns (\\r) and line feeds (\\n) from the beginning (^) and ending ($) of a string:

 s = s.replaceAll("(^[\\r\\n]+|[\\r\\n]+$)", "")

Complete Example:

public class RemoveLineBreaks {
    public static void main(String[] args) {
        var s = "\nHello world\nHello everyone\n";
        System.out.println("before: >"+s+"<");
        s = s.replaceAll("(^[\\r\\n]+|[\\r\\n]+$)", "");
        System.out.println("after: >"+s+"<");
    }
}

It outputs:

before: >
Hello world
Hello everyone
<
after: >Hello world
Hello everyone<

I'm going to add an answer to this as well because, while I had the same question, the provided answer did not suffice. Given some thought, I realized that this can be done very easily with a regular expression.

To remove newlines from the beginning:

// Trim left
String[] a = "\n\nfrom the beginning\n\n".split("^\\n+", 2);

System.out.println("-" + (a.length > 1 ? a[1] : a[0]) + "-");

and end of a string:

// Trim right
String z = "\n\nfrom the end\n\n";

System.out.println("-" + z.split("\\n+$", 2)[0] + "-");

I'm certain that this is not the most performance efficient way of trimming a string. But it does appear to be the cleanest and simplest way to inline such an operation.

Note that the same method can be done to trim any variation and combination of characters from either end as it's a simple regex.

Try this

function replaceNewLine(str) { 
  return str.replace(/[\n\r]/g, "");
}

String trimStartEnd = "\n TestString1 linebreak1\nlinebreak2\nlinebreak3\n TestString2 \n";
System.out.println("Original String : [" + trimStartEnd + "]");
System.out.println("-----------------------------");
System.out.println("Result String : [" + trimStartEnd.replaceAll("^(\\r\\n|[\\n\\x0B\\x0C\\r\\u0085\\u2028\\u2029])|(\\r\\n|[\\n\\x0B\\x0C\\r\\u0085\\u2028\\u2029])$", "") + "]");

1. Start of a string = ^ ,
2. End of a string = $ ,
3. regex combination = | ,
4. Linebreak = \r\n|[\n\x0B\x0C\r\u0085\u2028\u2029]

Another elegant solution.

String myString = "\nLogbasex\n";
myString = org.apache.commons.lang3.StringUtils.strip(myString, "\n");

For anyone else looking for answer to the question when dealing with different linebreaks:

string.replaceAll("(\n|\r|\r\n)$", ""); // Java 7
string.replaceAll("\\R$", "");          // Java 8

This should remove exactly the last line break and preserve all other whitespace from string and work with Unix (\n), Windows (\r\n) and old Mac (\r) line breaks: https://stackoverflow.com/a/20056634, https://stackoverflow.com/a/49791415. "\\R" is matcher introduced in Java 8 in Pattern class: https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html

This passes these tests:

// Windows:
value = "\r\n test \r\n value \r\n";
assertEquals("\r\n test \r\n value ", value.replaceAll("\\R$", ""));

// Unix:
value = "\n test \n value \n";
assertEquals("\n test \n value ", value.replaceAll("\\R$", ""));

// Old Mac:
value = "\r test \r value \r";
assertEquals("\r test \r value ", value.replaceAll("\\R$", ""));

String text = readFileAsString("textfile.txt");
text = text.replace("\n", "").replace("\r", "");