Replace preg_replace through preg_replace_callback

Question

I have a PHP library here and I want to replace preg_replace with preg_replace_callback.

This is the line:

preg_replace("/=([0-9A-F][0-9A-F])/e", 'chr(hexdec("$1"))', $l)

and this is what I did:

preg_replace_callback("/=([0-9A-F][0-9A-F])/", function($m) { return chr(hexdec($m["$1"])); }, $line);

But it didn't work. I still don't understand how preg_replace_callback works. I also looked into many other threads.

Can anyone help me please.

Thank you very much in advance!

Do a print_r of your `$m` array in your callback function. You would see that the structure is wrong. `$m["$1"]` should most likely be `$m[1]` — Jonathan Kuhn, Jun 13 '17 at 17:15

Wiktor Stribiżew · Accepted Answer · 2017-06-13T17:45:28.587

0

You need to use $m[1] to access the second element in the $m array:

preg_replace_callback("/=([0-9A-F][0-9A-F])/", function($m) { return  
    chr(hexdec($m[1])); }, $line
);

edited Jun 13 '17 at 17:45

answered Jun 13 '17 at 17:33

Wiktor Stribiżew

Why has your answer been marked as "community wiki"? Did you do that? or did someone else? Please explain if it's not too much trouble. (I just read a Meta post on community wiki, but I don't see the benefit.) – mickmackusa Jun 14 '17 at 05:35

1 Answers1