Error to parse the url value using regex

Question

I am parsing the url using regex. Some urls is not showing completely. I am using the following code.

it is working if the $string value is following

$string= "[video width="1280" height="720" mp4="http://egghead-site.s3.amazonaws.com/samples/DOS.mp4" title="DOS Software"][/video]" ;

if the mp4 value is following then it is not working. it is returning the half url only.

$string =[video width="1280" height="720" mp4="http://egghead-site.s3.amazonaws.com/samples/Colon%20Health%20Final%20V3.mp4" title="Synergi" ][/video]";  

$imageurl = !empty( $string ) ? preg_match_all('@((https?://)?([-\w]+\.[-\w\.]+)+\w(:\d+)?(/([-\w/_\.]*(\?\S+)?)?)*)@', $string , $match) : '';

return isset($match[0][0]) ? ut_portfolio_add_http($match[0][0]) : '';

I wanna to get the video url. I am new in regex so that I stuck at this point. Please let me know how can I get it in both cases.

Please could you write an example code without syntax errors. — Casimir et Hippolyte, Jul 28 '14 at 19:57
code is working but it shows the half url only http://egghead-site.s3.amazonaws.com/samples/Colon — user3885484, Jul 28 '14 at 20:05
@user3885484 No wonder, you have parts of it on a newline. Also your regex is far from matching an url. — HamZa, Jul 28 '14 at 20:07
possible duplicate of [What is the best regular expression to check if a string is a valid URL?](http://stackoverflow.com/questions/161738/what-is-the-best-regular-expression-to-check-if-a-string-is-a-valid-url) — HamZa, Jul 28 '14 at 20:08
This is a quick fix of your regex `((https?://)?([-\w%]+\.[-\w\.]+)+\w(:\d+)?(/([-\w/_\.]*(\?\S+)?)?)*)` Give it a try. — hex494D49, Jul 28 '14 at 20:08

score 0 · Answer 1 · answered Jul 28 '14 at 21:23

0

I just did a small change to your regex and it seems to be picking up exactly what you want. See here:

((https?:\/\/)+(.+\.[-\w\.]+)+\w(:\d+)?(/([-\w/_\.]*(\?\S+)?)?)*)

answered Jul 28 '14 at 21:23

IronWilliamCash

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Error to parse the url value using regex

1 Answers1