Hi Gerard, all the preg_* functions require delimiters surrounding regular expressions.
$foo = '\[this\](.*?)that'; should be by default: $foo = '/\[this\](.*?)that/'; the code you tried uses # as the delimiter instead of /, an option preg_* allows Take care, Greg -- phpDocumentor http://www.phpdoc.org "Gerard Samuel" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > The example doesn't have to make sense, but Im looking for the correct > syntax for $foo. I was trying -> > $foo = '\[this\](.*?)that'; > $bar = 'the other'; > > $str = preg_replace($foo, $bar, $other_string); > > But that doesn't work. I came across an example where the syntax of > $foo is in -> > $foo = '#\[this\](.*?)that#'; > > The second syntax of $foo works. I was wondering on the meaning of # in > the string?? > > Thanks > > -- > Gerard Samuel > http://www.trini0.org:81/ > http://dev.trini0.org:81/ > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
