RegEx in Java: how to deal with newline

Question

I am currently trying to learn how to use regular expressions so please bear with my simple question. For example, say I have an input file containing a bunch of links separ

Answer 1

The lines are probably separated by \r\n in your file. Both \r (carriage return) and \n (linefeed) are considered line-separator characters in Java regexes, and the . metacharacter won't match either of them. \s will match those characters, so it consumes the \r, but that leaves .* to match the \n, which fails. Your tester probably used just \n to separate the lines, which was consumed by \s.

If I'm right, changing the \s to \s+ or [\r\n]+ should get it to work. That's probably all you need to do in this case, but sometimes you have to match exactly one line separator, or at least keep track of how many you're matching. In that case you need a regex that matches exactly one of any of the three most common line separator types: \r\n (Windows/DOS), \n (Unix/Linus/OSX) and \r (older Macs). Either of these will do:

\r\n|[\r\n]

\r\n|\n|\r

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Update: As of Java 8 we have another option, \R. It matches any line separator, including not just \r\n, but several others as defined by the Unicode standard. It's equivalent to this:

\r\n|[\n\x0B\x0C\r\u0085\u2028\u2029]

Here's how you might use it:

(?im)^.*www.*\R.*Pig.*$

The i option makes it case-insensitive, and the m puts it in multiline mode, allowing ^ and $ to match at line boundaries.