Greedy with a trailing greedy Regex in javascript

Question

I am trying to get "a=.*" from the argument list in a URL (http:/a.b.com/?a=123) in Javascript.

/(.*[&|?])(a=.+?)(&.*)/.exec ("http:/a.b.com/?a=123&b=123")

gives me the expected and desired :

["http:/a.b.com/?a=123&b=123", "http:/a.b.com/?", "a=123", "&b=123"]

However, there might not be a following "&arg=val" and I need to be able to search for an optional trailing "&.*" (so adding '?' after the '&' on the third group. Hence :

/(.*[&|?])(a=.+?)(&?.*)/.exec ("http:/a.b.com/?a=123&b=123")

which gives me the unexpected :

["http:/a.b.com/?a=123&b=123", "http:/a.b.com/?", "a=1", "23&b=123"]

The second group now has only 1 character after the '=' and the remainder two '23' are part of the third group..

What I might be doing wrong ? Also any advice to do this a different/better way is also welcomed.

Thanks very much in advance !

Lucas Trzesniewski · Accepted Answer · 2014-10-11 21:42:18Z

1

You have two possibilities:

[&?]a=(.+?)(?:&|$)
[&?]a=([^&]+)

So, either match non-greedily until you encounter a & or the end of the string, or match greedily all characters except &.

Also, I've shortened your regex and made the group only capture the value of the a parameter, without the a= part. The .* are not needed, your regex doesn't have to match the whole string unless it's anchored with ^ and/or $. And that | in the character class was also wrong.

Example with your test case:

alert(JSON.stringify(
  /[&?]a=([^&]+)/.exec("http:/a.b.com/?a=123&b=123")
))

answered Oct 11, 2014 at 21:42

Lucas Trzesniewski

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5 Comments

user1055761 Over a year ago

I can understand the second option and used it and worked !. I am not familiar with ?: - so just going with the second option. (thanks for pointer on ?: - will check it out). Great stuff Lucas and this was a lightening fast response !

Lucas Trzesniewski Over a year ago

You're welcome. FYI, (?:...) is equivalent to (...) except it doesn't capture it's value. It's only used for simple grouping.

user1055761 Over a year ago

Oh, so one could also do [&?]a=(.+?)[&|$]+ ?

Lucas Trzesniewski Over a year ago

No, you have to use (&|$) because in a character class ([...]), regex metacharacters lose their special meaning. So, [&|$] means: one of the following characters: & or | or $ literally.

user1055761 Over a year ago

Used some other bits from your enhanced answer. I need to get the remainder of the URL hence the '.*'. I wrapped it with ^ and $ as per your recommendation. Alls working well. Thanks !

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