开发者

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

开发者 https://www.devze.com 2023-01-03 07:05 出处:网络
I have this URL: http://example.com/createSend/step4_1.aspx?cID开发者_JS百科=876XYZ964D293CF&snap=true&jlkj=kjhkjh&

I have this URL:

http://example.com/createSend/step4_1.aspx?cID开发者_JS百科=876XYZ964D293CF&snap=true&jlkj=kjhkjh&

And this regex pattern:

cID=[^&]*

Which produces this result:

cID=87B6XYZ964D293CF

How do I REMOVE the "cID="?

Thanks


You can either use lookbehind (not in Javascript):

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

Or you can use grouping and grab the first group:

cID=([^&]*)


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

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


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


By using capturing groups:

cID=([^&]*)

and then get $1:

87B6XYZ964D293CF


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]);


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];
0

精彩评论

暂无评论...
验证码 换一张
取 消

关注公众号