Regex

[richie]

Kindly help write regex to replace the following pattern

part a [img:hf]http://mydom.net/images/redface.gif[/img:hf] part b

In the end, I want to be left with:

part a part b

I tried the following, but it came back blank...

$message = 'part a part b';

$img_pattern = '[img(.)](.)?[\/img(.)]';

$message = preg_replace($img_pattern,'', $message);

echo $message;

[mattm591]

Well this doesn't use preg_replace but gives the effect you desired (as far as I understood it),

<?php

$message = 'part a http://mydom.net/images/redface.gif part b';

$a = strpos($message,"http");
$b = strpos($message,"gif");
$c = $b + 3;

$tmp = substr($message,0,$a);
$tmp2 = substr($message,$c);

$newmessage = $tmp . $tmp2;

echo $newmessage;

?>

If you wanted to get rid of the ");
$c = $b + 6;[/code]

[halojoy]

I got it working in first attempt!!!!!!

What I did was this:
we have 3 parts:
[img:hf]
.*
[/img:hf]

They should all 3 be removed. So we put (....) around it all.
- preg_replace removes the match, (....)
while
- preg_match keeps the match, (....)

Now, the following will work if there is only 1 such match in a string.
I do not know what will happen if there are several matches in a string ...

One little thing.
Result will be

'part a  part b'
*notice: there are 2 spaces between part a and part b,
just as expected.

Could be converted to 1 space, using:

str_replace( '  ', ' ', $message);

my code is this:

<?php

$msg = 'part a [img:hf]http://mydom.net/images/redface.gif[/img:hf] part b';

$pattern = '#(\[img:hf\].*\[/img:hf\])#';

$message = preg_replace($pattern, '', $msg);

// test before and after
echo $msg;
echo '<br>';
echo $message;

regars 🙂
halojoy