How can I match on, but exclude a regex pattern?

Question

I have this URL:

http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&

And this regex pattern:

cID=[^&]*

Which produces this result:

cID=87B6XYZ964D293CF

How do I REMOVE the "cID="?

Thanks

Elytron

what language? are you working in? – Besetting

Which regex parser are you using? – Bobinette

Sorry, I should have been more specific. I'm working in javascript – Elytron

Answer 1

You can either use lookbehind (not in Javascript):

(?<=cID=)[^&]*

Or you can use grouping and grab the first group:

cID=([^&]*)

Heliopolis

Answer 2

Generally speaking, to accomplish something like this, you have at least 3 options:

• Use lookarounds, so you can match precisely what you want to capture
  • No lookbehind in Javascript, unfortunately
• Use capturing group to capture specific strings
  • Near universally supported in all flavors
• If all else fails, you can always just take a substring of the match
  • Works well if the length of the prefix/suffix to chop is a known constant

References

• w3schools - jsref - substring
• regular-expressions.info/Capturing groups and Lookarounds
  • Flavor comparison

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Examples

Given this test string:

i have 35 dogs, 16 cats and 10 elephants

These are the matches of some regex patterns:

• \d+ cats -> 16 cats (see on rubular.com)
• \d+(?= cats) -> 16 (see on rubular.com)
• (\d+) cats -> 16 cats (see on rubular.com)
  • Group 1 captures 16

You can also do multiple captures, for example:

• (\d+) (cats|dogs) yields 2 match results (see on rubular.com)
  • Result 1: 35 dogs
    • Group 1 captures 35
    • Group 2 captures dogs
  • Result 2: 16 cats
    • Group 1 captures 16
    • Group 2 captures cats

Epode

Answer 3

With JavaScript, you'll want to use a capture group (put the part you want to capture inside ()) in your regular expression

var url = 'http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&';

var match = url.match(/cID=([^&]*)/);
// ["cID=876XYZ964D293CF", "876XYZ964D293CF"]

// match[0] is the whole pattern
// match[1] is the first capture group - ([^&]*)
// match will be 'false' if the match failed entirely

Agrology

Answer 4

By using capturing groups:

cID=([^&]*)

and then get $1:

87B6XYZ964D293CF

Fairminded

Answer 5

Here's the Javascript code:

 var str = "http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&";
    var myReg = new RegExp("cID=([^&]*)", "i");
    var myMatch = myReg.exec(str);
    alert(myMatch[1]);

Varick

Answer 6

There is a special syntax in javascript which allows you to exclude unwanted match from the result. The syntax is "?:" In your case the solution would be the following

'http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&'.match(/(?:cID=+)([^&]*)/)[1];

Premature

This is excessive. 'http://example.com/createSend/step4_1.aspx?cID=876XYZ964D293CF&snap=true&jlkj=kjhkjh&'.match(/cID=+([^&]*)/)[1]; is simpler as it does not require the non-capturing group and returns the same result as your example. – Urson