Java regex: newline + white space

Question

should be simple, but I'm going crazy with it.

Given a text like:

line number 1
line number 2
 line number 2A
line number 3
 line number 3A
 line number 3B
line number 4

I need the Java regex that deletes the line terminators then the new line begin with space, so that the sample text above become:

line number 1
line number 2line number 2A
line number 3line number 3Aline number 3B
line number 4

Strathspey

Try "\r?\n\W+". I haven't tried it, so I don't want to register it as an answer. In case you haven't checked it out already, you can find some specifics about Java regex here: download.oracle.com/javase/6/docs/api/java/util/regex/… ... If you are guaranteed that "line" is the first text on a given line, you could also try "\r?\n[^l]+". – Aliform

@Mike M: Won't work on Apple's OSs – Outbalance

Should work on OS-X though, right? Are people still using old versions of Apple? – Aliform

Answer 1

yourString.replaceAll("\n ", " "); this wont help?

Alamo

1) The question says: _if you encounter line that begins with <blank> remove both the blank and the new line. 2) '\n' is not portable – Outbalance

@MatteoSp, remeber that Stings are immutable, so in fact you need yourString = yourString.replaceAll("\n ", " "); or something like that, because calling method wont change String object itself permanently. Glad to help :) – Alamo

Answer 2

String res = orig.replaceAll("[\\r\\n]+\\s", "");

Outbalance

Answer 3

Perhaps to make it cross-platform:

String pattern = System.getProperty("line.separator") + " ";
string.replaceAll(pattern, "");

Uncover

Well, the system line.separator doesn't really help, since you don't know if the files were created on the system where the JVM is running – Outbalance

Answer 4

"\n " This is should do the trick if you are in Unix LF mode. For DOS like you need to match CRLF "\r\n ". Did check with RegexBuddy looking fine.

Crackpot