Java trim new lines

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]) + "-"); 

// 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.

Yeah but what if you don't know how many lines are at the beginning/end? Your solution assumes there are exactly 2 newlines in both cases

The second parameter of split() is just the limit. Leave it off if you want to match an unlimited number of times.

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]

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

// 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$", "")); 

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

In Java, you may come across a scenario where you need to remove newline characters from the beginning and end of a string. This can be important when working with text-based data and you want to eliminate any unnecessary white spaces or line breaks. The newline characters, represented by \n, can cause issues when processing data or presenting it in a certain format. In this tutorial, we will explore different methods to remove newline characters from the start and end of a string in Java.

Method 1: Using String.trim() Method

To remove newlines from the beginning and end of a string in Java, you can use the String.trim() method. This method returns a copy of the string with leading and trailing whitespace removed, including newlines. Here's an example:

String str = "\n\nHello, world!\n\n"; String trimmed = str.trim(); System.out.println(trimmed);

In this example, we create a string str with newlines at the beginning and end. We then use the trim() method to remove the newlines and any other whitespace from the beginning and end of the string. Finally, we print the trimmed string to the console.

You can also use trim() to remove newlines from the beginning and end of a string array. Here's an example:

String[] arr = "\n\nHello", "world!\n\n">; for (String s : arr)  System.out.println(s.trim()); >

In this example, we create a string array arr with newlines at the beginning and end of each element. We then use a for loop to iterate over the array and print each element with the newlines removed using the trim() method.

That's it! Using String.trim() is a simple and effective way to remove newlines from the beginning and end of a string in Java.

Method 2: Using Regular Expressions

To remove newlines from the beginning and end of a string in Java using regular expressions, you can use the replaceAll() method with the regular expression ^\n+|\n+$ as the first argument and an empty string as the second argument. Here's an example code:

String str = "\n\nHello, world!\n\n"; String trimmedStr = str.replaceAll("^\n+|\n+$", ""); System.out.println(trimmedStr); // Output: "Hello, world!"

• ^ : Matches the start of the string.
• \n+ : Matches one or more newline characters.
• | : Or operator.
• $ : Matches the end of the string.

So the regular expression ^\n+|\n+$ matches one or more newline characters at the beginning of the string or at the end of the string.

The replaceAll() method replaces all occurrences of the regular expression with the second argument, which is an empty string in this case.

Note that the regular expression ^\n+|\n+$ can also be written as ^\s+|\s+$ to remove any whitespace characters (including newlines) from the beginning and end of the string.

Here's another example code that demonstrates how to remove newlines from a list of strings using a regular expression pattern:

ListString> lines = Arrays.asList("Hello\n", "\nworld\n", "!\n\n"); Pattern pattern = Pattern.compile("^\n+|\n+$"); ListString> trimmedLines = lines.stream() .map(pattern::matcher) .map(m -> m.replaceAll("")) .collect(Collectors.toList()); System.out.println(trimmedLines); // Output: ["Hello", "world", "!"]

This code uses the Pattern class to compile the regular expression pattern ^\n+|\n+$ , and then uses the Matcher class to apply the pattern to each string in the list. The map() method is used to apply the replaceAll() method to each Matcher object, and the collect() method is used to collect the results into a new list.

Method 3: Using a Custom Function

To remove newlines from the beginning and end of a Java string using a custom function, you can use the following code:

public static String removeNewlines(String input)  int start = 0; int end = input.length() - 1; while (start  end && input.charAt(start) == '\n')  start++; > while (end >= start && input.charAt(end) == '\n')  end--; > return input.substring(start, end + 1); >

This function takes a string as input and returns a new string with all newlines removed from the beginning and end of the input string. The function first finds the start and end indices of the substring that contains the non-newline characters by iterating over the string from both ends and skipping over newlines. Finally, it returns the substring between the start and end indices.

Here are some examples of how to use this function:

String input1 = "\n\nhello world\n"; String output1 = removeNewlines(input1); // output1 is "hello world" String input2 = "\n\n\n"; String output2 = removeNewlines(input2); // output2 is "" String input3 = "no newlines"; String output3 = removeNewlines(input3); // output3 is "no newlines"

In the first example, the input string contains newlines at the beginning and end, which are removed by the function. In the second example, the input string consists only of newlines, which are all removed. In the third example, the input string contains no newlines, so the function returns the original string unchanged.

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Removing new lines/spaces from beginning and end of string Java

I have a problem - I can't seem to be able to remove the new lines/spaces from the beginning/end of a string. I use \s in the beginning and end of the regex and even use .trim() after I get the string, but to no avail.

public void extractInfo(String mydata) < // regex to extract the user name Pattern pattern = Pattern.compile("user:\\s*(.*)\\s+branch"); Matcher matcher = pattern.matcher(mydata); // regex to extract the branch name Pattern pattern2 = Pattern.compile("branch:\\s*(.*)\\s+changed"); Matcher matcher2 = pattern2.matcher(mydata); // regex to extract the comment and write it in a variable comment = mydata.replaceAll("(?s)\\s.*java;[0-9,.]+|.*java;NONE\\s", ""); // put the author name in a variable matcher.find(); author = matcher.group(1).toString(); // put the branch name in a variable matcher2.find(); branch = matcher2.group(1).toString(); author.trim(); comment.trim(); branch.trim(); >

This is what I use to extract the info. This is the output I get (lines kept), after I append the extracted information using StringBuilder:

git log --all -100 --before="2013-03-11" --branches=HEAD --author="\(cholakov\)" --grep="^[#]*[0]*23922:[ ]*user: 

The new line after user: is what causes the whole command to fail when I try to execute it in cmd, that's what I need fixed. And this is my input (can't seem to be able to keep the formatting, DataObjectParser.java;1.94 is on a new line and there is no line skipped between each line):

user: cholakov branch: HEAD changed files: DataObjectParser.java;1.94 Fixed the message for defaulted bonds

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