I need to remove everything around the link below. I'm using preg_match with regular expression, and since I'm new to the area, I'm having trouble finding my error. Could you please help me out?
How do I do:
preg_match('/www.facebook.com/plugins/video.php?href=\/(.+)&show_text=0&width=560', $v, $output
Full link: link
I need to: "https% 3A% 2F% 2Fwww.facebook.com% 2Fportateste% 2Fvideos% 2F1585186298555265% 2F"
Thank you
You can use the following expression:
href=(.*?)&
In this way it will capture everything from href to first & .
Example:
<?php
preg_match("/href=(.*?)&/", $link, $result);
var_dump( $result[1] );
If you want something more complex, just use the expression below:
href=(.*?)(?(?=&)&|$)
└─┬─┘ └┬─┘└┬┘└┬┘
│ │ │ └── Caso não haja, selecione até o último caractere.
│ │ └───── Caso haja, seleciona até ele.
│ └───────── Verifique se há '&' na URL
└────────────── Captura tudo após o 'href='