I love preg_mach_all
command, but now I have this situation:
$lotes_fil='/((?:https?|http?):\/\/(?:\S*?\.\S*?)(?:[\s)\[\]{};"\'<]|\.\s|$))/';
preg_match_all($lotes_fil,$rs,$link_pre, PREG_SET_ORDER);
The output is this:
[0] => Array
(
[0] => https://s3.amazonaws.com/mesena.com/vodcategories/58e1a299dca3c3a22fbd8319/iconPng-1576021669572.png"
[1] => https://s3.amazonaws.com/mesena.com/vodcategories/58e1a299dca3c3a22fbd8319/iconPng-1576021669572.png"
)
[1] => Array
(
[0] => https://images.mesena.com/assets/images/default/vodcategory.id-imageFeatured.jpg"
[1] => https://images.mesena.com/assets/images/default/vodcategory.id-imageFeatured.jpg"
)
[2] => Array
(
[0] => https://s3.amazonaws.com/silo.mesena.com/origin/5638f208fbdc021398ae8681/production/202006/19/5638f208fbdc021398ae8681_212-21719"
[1] => https://s3.amazonaws.com/silo.mesena.com/origin/5638f208fbdc021398ae8681/production/202006/19/5638f208fbdc021398ae8681_212-21719"
)
I need eliminate the last " but i want use this regex beacuse working for me well.In the original data have " as the ending for each URL
Thx for your help.
I'd use:
~https?://[^\s"]+~
Explanation:
~ # regex delimiter, it allows to use / without escaping
https? # http OR https
:// # literally
[^\s"]+ # 1 or more any character that is not a space or double quote