Question

Hello I need to find regex that would get middle section of this text:

# Command // first line with symbol character 

First line of interest
Second line of interest
\n
Third line of interest
\n
\n // I am not interested in trailing new lines.

How could I get text starting with First line of interest and ending wint third line of interest? Thank you.

Answer 1

String test = "# Command\n\nFirst line of interest\r\nSecond line of interest\n\r\nThird line of interest\r\n\n";
System.out.printf("%n>>%s<<%n", test);

Pattern p = Pattern.compile("^(?!#).+(?:[\r\n]+.+)*", Pattern.MULTILINE);
Matcher m = p.matcher(test);
if (m.find())
{
  System.out.printf("%n>>%s<<%n", m.group());
}

output:

>># Command

First line of interest
Second line of interest

Third line of interest

<<

 

>>First line of interest
Second line of interest

Third line of interest<<

---

The match starts at the beginning (^ in MULTILINE mode) of the first line that does not start with a hash symbol ((?!#)), but does contain characters other than line separators (.+, not .*).

[\r\n]+ matches one or more line separators, whether they be the Unix (\n), DOS (\r\n), or older-Mac (\r) style of separator. You should always be prepared to handle any or all of the different line separators, no matter what platform your code is running on.

(?:[\r\n]+.+)*, then, matches zero or more additional lines, without matching any trailing line separators.