Given /foo/bar/image.jpg?x=1&y=2, how do I obtain image.jpg?
https://stackoverflow.com/a/423385 provides a partial answer, but does not address the GET parameters.
4 Answers
You can use regex as others have suggested, but I find this more readable:
var src = '/foo/bar/image.jpg?x=1&y=2';
var img = src.split('/').pop().split('?')[0];
console.log(img);
1 Comment
user1032531
All answers work, and I can't tell which one is best, but this one is a couple of characters shorter than the rest and maybe more readable (at least for a regex hack like me!).
Since the ? symbol separates the GET parameters from the URL, you can
str=str.split("?")[0]
filename = str.replace(/^.*[\/]/, '')
Comments
You can use this regex:
/\/([^?\/]+(?=\?|$))/
and use captured grpup #1.
/ # matches a literal /
[^?\/]+ # matches 1 or more of any char that is not ? or /
(?=\?|$) # is a lookahead to assert that next position is ? or end of line
2 Comments
user1032531
Includes a leading forward-slash. Thanks for the explanation,however!
anubhava
NO it doesn't include
/ in captured group. I wrote clearly use captured grpup #1 in my answer.Try:
var str = "/foo/bar/image.jpg?x=1&y=2";
var fileName = str.split('/').slice(-1)[0].split('?')[0];
or, for a regex method:
var fileName = str.split('/').slice(-1)[0].match(/[^?]+/)[0];
/foo/bar/image.jpg?x=1&y=2is thesrcof an image, and not the page URL. Does your recommendation still apply?