Java Regular Expression Escape Sequence

Question

I was trying to match the example in , <p><a href="example/index.html">LinkToPage</a></p>

With rubular.com I could get something like <a href=\"(.*)?\/index.html\">.*<\/a>.

I'll be using this in Pattern.compile in Java. I know that \ has to be escaped as well, and I've come up with <a href=\\\"(.*)?\\\/index.html\\\">.*<\\\/a> and a few more variations but I'm getting it wrong. I tested on regexplanet. Can anyone help me with this?

No escaping necessary here. Just use ...=\"... A backslash only needs to be escaped, when you actually want a backslash. And in a regex, you have to do it twice. — jlordo
– jlordo, Commented Jun 3, 2013 at 19:34
If it's an escaping issue, try printing your string to the command line to figure out what it thinks it is and correcting accordingly. All those backslashes can get annoying. — chessbot
– chessbot, Commented Jun 3, 2013 at 19:35
Since this is HTML, you should consider using an HTML parser... Like, for instance, jsoup. — fge
– fge, Commented Jun 3, 2013 at 19:38

Laurence Gonsalves · Accepted Answer · 2013-06-03 19:43:15Z

2

Use "<a href=\"(.*)/index.html\">.*</a>" in your Java code.

You only need to escape " because it's a Java string literal.

You don't need to escape /, because you aren't delimiting your regex with slashes (as you would be in Ruby).

Also, (.*)? makes no sense. Just use (.*). * can already match "nothing", so there's no point in having the ?.

answered Jun 3, 2013 at 19:43

Laurence Gonsalves

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Community · Accepted Answer · 2017-05-23 11:43:37Z

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Pattern.compile("<a href=\"(.*)?/index.html\">.*</a>");

That should fix your regex. You do not need to escape the forward slashes.

However I am obligated to present you with the standard caution against parsing HTML with regex:

RegEx match open tags except XHTML self-contained tags

edited May 23, 2017 at 11:43

CommunityBot

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answered Jun 3, 2013 at 19:44

Aurand

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John Humphreys · Accepted Answer · 2015-04-15 21:12:17Z

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You can tell Java what to match and call Pattern.quote(str) to make it escape the correct things for you.

answered Apr 15, 2015 at 21:12

John Humphreys

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