Grabbing the third fragment between square brackets

Still completely stuck with regex's and square brackets. Hopefully someone can help me out.

Say I have a string like this:

room_request[1][1][2011-08-21]

How would I grab the third fragment out of it?

I tried the following, but I'm not exactly sure what I'm doing so it's fairly hard to figure out where I'm going wrong.

.match(/\[(.*?)\]/);

But this returns the [1] fragment. (The first one, I guess). So then, I asked here on SO 开发者_JAVA百科and people told me to add a global flag:

.match(/\[(.*?)\]/g)[2];

In other cases that I've used this regex, this worked fine. However, in this case, I want the stuff INSIDE the square brackets. It returns:

[2011-08-21]

But I really want 2011-08-21.

How can I do this? Thanks a lot.

If anyone could recommend any decent resources about regular expressions, that'd be great aswell. I'm starting to understand the very basics but most of this stuff is far too confusing atm. Thanks.

Two possible methods. To grab the third bracketed expression:

.match(/\[.*?\]\[.*?\]\[(.*?)\]/);

Or, if you know that the expression you want is always at the end of the string:

.match(/\[(.*?)\]$/);

var str = "room_request[1][1][2011-08-21]"
var val = str.match(/\[[^\]]*\]\[[^\]]*\]\[([^\]]*)\]/);
alert(val[1]);

This is a little less messy I think:

var r = "room_request[1][1][2011-08-21]";
var match = r.match(/(?:\[([^\]]+)\]){3}/);
console.log(match[1]);

Basically, it picks out the third match of the square brackets containing something. You get the match result back with two matches - the whole [1][1][2011-08-21] (for whatever reason) and the matched date: 2011-08-21

My regex is a little rusty, but this certainly works.