i can't taking a part from href value with preg_match_all

Question

is there any way to extract a part from href value? For example my code is :

$searchfrom = '@<a class="uye" href="index.php?profil=12" contenteditable="false">@';
$search = '@<a class="uye" href="index.php?profil=(.*?)" contenteditable="false">@';
preg_match_all($search,$searchfrom,$sonuc); 

i want to extract 12 from searchfrom variable.

Answer 1

You have made 3 errors but your pattern doesn't work for only one reason.

How to write a literal string in a pattern?

? is a special character in a regex and means optional (after a character) or turn a quantifier to lazy (after +, *, ?, {n,m}) or is part of syntax for several features (non capturing or atomic groups, conditionals, recursion, definitions).
If you want to write it as literal in a pattern, you must escape it:

php?profil => php\?profil

This is the only reason your pattern doesn't work.

The same with a literal dot. . has a special meaning and must be escaped:

index.php => index\.php

But since . means all characters, it matches the literal dot too.

There are twelve special characters that must be escaped in a pattern to be used as literal:

[ ( ) { ? * + ^ $ \ | .

Note that when the opening curly bracket { is not followed by something like this: 2,5} or ,5} or 2,}, the situation isn't ambiguous (i.e. the opening curly bracket is not the begining of a {n,m} quantifier.) and you don't need to escape it.

However an other character must be escaped: the pattern delimiter.

With PHP you must use a pattern delimiter to enclose a pattern, the most used is the slash / but you can use an other character like ~ # @ ... (note that using a regex special character or a character that is used as literal many times in the pattern is a bad idea.)

You seems to have forgotten to put the delimiters since the @ are part of the original string. In the pattern, the @ is seen as a delimiter because it is the first character, but it is not seen as a literal @. If you want to match the @ you must enclose your pattern with delimiters.

So, the correct pattern is :

$search = '/@<a class="uye" href="index\.php\?profil=(.*?)" contenteditable="false">@/';

If you want to avoid to search and escape each special character in a literal string, you can enclose your literal string between \Q....\E:

$search = '/\Q@<a class="uye" href="index.php?profil=\E(.*?)\Q" contenteditable="false">@\E/';

(the pattern delimiter (if present), must always be escaped.)

Or you can use the preg_quote() function:

$search = '/' . preg_quote('@<a class="uye" href="index.php?profil=', '/') . '(.*?)' . preg_quote('" contenteditable="false">@', '/') .'/';

The second (optional) argument of preg_quote() is the delimiter. I have written it, but it is / by default.