Re: Regular Expression Query
| From: | Cybercandy | Date: | Sat, 06 Jan 2001 19:20:23 +0000 |
| Subject: | Re: Regular Expression Query | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-33151@lists.php.net to get a copy of this message | ||
Thanks for those two points. I still haven't cracked this.
I'm trying to break the problem down so I've tried the following
example:
$semifile2 = ereg_replace("\#39","YES",$semifile);
But where there is a string: #39 it doesn't replace it with the
word YES, however if I remove \# then it does replace
39 with YES
Thanks for the help so far.
Allan
www.cybercandy.co.uk
----- Original Message -----
From: "Rasmus Lerdorf" <rasmus@php.net>
To: "Cybercandy Ltd" <allan@cybercandy.co.uk>
Cc: <php-general@lists.php.net>
Sent: Saturday, January 06, 2001 6:53 PM
Subject: Re: [PHP] Regular Expression Query
> > I think the most difficult thing I've encountered so far in trying to
> > learn PHP, as a novice programmer, is regular expressions.
> >
> > I'm trying to use ereg_replace on a string containing a large quantity
> > of data taken from a webpage which I then parse using
> > the explode function.
> >
> > One regularly occuring string is: (Reference #xxx)
> >
> > Where xxx is a number of up to 3 digits.
> >
> > I want to replace this string using back references to just xxx
> > and I've tried using the following:
> >
> > $out = ereg_replace("\(Reference #.([0-9]+).\)","\\1",$in);
> >
> > I just can't suss why this doesn't work.
> >
> > Any pointers anyone?
>
>
> What are those .'s for in your regex?
>
> Just make it: "\(Reference #([0-9]+)\)"
> and I bet it will work.
>
> with the .'s there you would only get the middle digit of your xxx
>
> -Rasmus
>
>
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