Java正则表达式转义序列

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?

Answer 1

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 ? .

Answer 2

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

Answer 3

您可以告诉Java匹配什么，然后调用Pattern.quote（str）使其逃避正确的事情。