Here is an good example for the variable holding the pattern ->
$foo = '/\[url\]([a-z]+://.*?)\[/url\]/';
This pattern would not work, but if I change it to
$foo = '#\[url\]([a-z]+://.*?)\[/url\]#';
It does work.
Not complaining but just trying to figure out why the first version doesn't work for future references.
Thanks
Greg Beaver wrote:
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
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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/
-- Gerard Samuel http://www.trini0.org:81/ http://dev.trini0.org:81/
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