Removing a hard Enter from a String

Question

What is the best way to remove a hard enter from a String?

Input:

String in= "strengthened columns 
with GRPES
";

Expected output: strengthened columns with GRPES

I tried the below code, but it's not working for me.

in = in.replaceAll("\\r\\n","");
System.out.println(in);

• M. Prokhorov: \r\n is not the only possible variant. It could be just one \n char in there. But also - if you aren't getting the correct output - what DO you get?
• Mahesh M: @M.Prokhorov, I tried, but I'm not getting the correct output.
• Mark Rotteveel: In the regex, you should use \r\n, not \\r\\n (though \\r\\n should work as well). Please show the actual codepoints of the string.
• g00se: in = in.replaceAll("[\r\n]","");
• g00se: OK will add as answer for you to accept

Answer 1

Unless you don't have a specific reason to use java-7 today, Here's a solution using java 13 or above

 String in= """
 strengthened columns 
 with GRPES
 """;
 in = in.replaceAll("\\n","");
 System.out.println(in);

I have observed the question is tagged with java-7, do let me know if you are looking for a solution specific to the version

• Mahesh M: thank you for answering. I am currently using Java 7, so I tagged it as Java 7. I tried the code below (in = in.replaceAll("[\r\n]","");) and it is working fine.

Answer 2

Actually you don't escape standard escape sequences when you use regexes. Also you don't want to specify an order of escape sequences - you just want to eliminate any type of line separator, so

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

With later versions of Java, that could probably be

in = in.replaceAll("\\R","");