How to write regexp to get a paramater from URL

Question

I need help with a regexp to get only the code parameter from this url:

URL: code=4%2FsAB_thYuaw3b12R0eLklKlc-qcvNg6f8E8pgvu_02MTTJE0NOcyvXkrTQrB3QK8209wSbIpLDEBrk8vUGYKQ41I&scope=https:.....

My Code:

getParameterByName(name) {
 var url = window.location.href;
 name = name.replace(/[\[\]]/g, '\\$&');
 var regex = new RegExp('[?&]' + name + '(=([^&#]*)|&|#|$|)'),
 results = regex.exec(url);
 if (!results) return null;
 if (!results[2]) return '';
 return decodeURIComponent(results[2].replace(/\+/g, ' '));
}

this piece of code is returning `

4/sAB_thYuaw3b12R0eLklKlc-qcvNg6f8E8pgvu_02MTTJE0NOcyvXkrTQrB3QK8209wSbIpLDEBrk8vUGYKQ41I

In this case i need the first and only backslash to be a % sign instead..

I can't get it to work.

Thnx in advance!

• Mark: Not sure I fully understand what you are trying to do but if you just want everything between "code=" and the next "&" try "code=([^&]*)" I assume you have a reason for not doing this, if so, sorry.

Answer 1

It works for me!!!

function getUrlParameterByName(name) {
 name = name.replace(/[\[]/, "\\[").replace(/[\]]/, "\\]");
 var regex = new RegExp("[\\?&]" + name + "=([^&#]*)"),
 results = regex.exec(location.search);
 return results === null ? "" : decodeURIComponent(results[1].replace(/\+/g, " "));
}

• Luis Estevez: I like this because you answered it in the format of KENUBBE despite the fact KENUBBE can do this with one regex without using replace or any other operation.
• Mark: That's why I was confused by the question - his code is so overly complicated I thought I must be missing something.

Answer 2

Untested JavaScript regex:

var myRe = /=(.*)&/igm
var url = 'code=4%2FsAB_thYuaw3b12R0eLklKlc-qcvNg6f8E8pgvu_02MTTJE0NOcyvXkrTQrB3QK8209wSbIpLDEBrk8vUGYKQ41I&scope=https:';
var code = myRe.exec(url);

Answer 3

Use \bcode=([^&]+). It's sure to grab code query using word boundary 'code=' then capture any characters until it reaches & or the end.

test regex here: https://regex101.com/r/eSnB4S/1

• Mark: Does the {1,} add anything to just * ?
• Luis Estevez: @Mark yes. {1,} makes sure it matches between one and unlimited. * can match between zero and unlimited. So if you use * quantifier instead, and have test case code= you will have an empty string in capture group. So having {1,} helps prevent that empty string.
• Code Maniac: @LuisEstevez well + does the same thing {1,}. so you can use + if you want to match one or more. demo regex101.com/r/eSnB4S/2
• Luis Estevez: @CodeManiac Ahh, yes thanks! I'll use that for now on :)
• Mark: I know that, I'm just looking for a real world case where that matters here. The only difference would be where code=&... my version produces an empty match group and yours produces no match. But that is also a malformed url with queries so...