Re: [PHP] quick question about preg_replace
On Thu, 05 Oct 2006 09:29:46 +0800, Penthexquadium wrote:
On Wed, 4 Oct 2006 23:19:58 +0200, Martin Bach Nielsen [EMAIL PROTECTED]
wrote:
Hi all.
I have written a guestbook (I know, there's a ton out there), which until
recently did not get any spam, so now I'm trying to remove the spam-ad by
using preg_replace:
$melding = preg_replace('/(a href=\http:\/\/[^^]+)/i', This message is
temporarily cut-off due to spam-suspicion., $melding);
However, this only makes the problem half-solved, since I also want any text
before the link to be replaced with a message as stated above. The above
line removes anything after 'a href=//'.
Is there any way to make the above line to include anything before the a
href=// ?
I have tried different options, and yes, I have read the php manual on
preg_replace, but I might not have properly identified how to get the text
in front modified.
Thankful for any hints, tips, links, anything that helps :-)
Regards,
Martin Bach Nielsen
If you only want to replace the line including url(s) by a message, a
simple regular expression is enough.
?php
$melding = This is a spam.\n
. This is a a href=\http://www.example.com/\;example/a line.\n
. Another line.\n
. a href=\http://www.e.com/ads/show?n=123\;123/a (456)\n
. End Line.\n;
$melding = preg_replace(/.*?a href=\http:\/\/.+/i, This line is
temporarily cut-off due to spam-suspicion., $melding);
echo $melding;
?
The script above will output:
This is a spam.
This line is temporarily cut-off due to spam-suspicion.
Another line.
This line is temporarily cut-off due to spam-suspicion.
End Line.
If you use this code, be sure to use lt; in stead of and gt; in stead
of or else the spam messages won't be shown.
Otherwise, if you want the entire message changed, it's quicker to do:
if (preg_match('/(a href=\http:\/\/[^^]+)/i', $melding)) {
$melding = 'This message is temporarily cut-off due to spam-suspicion.';
}
Just an idea...
Ivo
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[PHP] quick question about preg_replace
Hi all.
I have written a guestbook (I know, there's a ton out there), which until
recently did not get any spam, so now I'm trying to remove the spam-ad by
using preg_replace:
$melding = preg_replace('/(a href=\http:\/\/[^^]+)/i', This message is
temporarily cut-off due to spam-suspicion., $melding);
However, this only makes the problem half-solved, since I also want any text
before the link to be replaced with a message as stated above. The above
line removes anything after 'a href=//'.
Is there any way to make the above line to include anything before the a
href=// ?
I have tried different options, and yes, I have read the php manual on
preg_replace, but I might not have properly identified how to get the text
in front modified.
Thankful for any hints, tips, links, anything that helps :-)
Regards,
Martin Bach Nielsen
Re: [PHP] quick question about preg_replace
On Wed, 4 Oct 2006 23:19:58 +0200, Martin Bach Nielsen [EMAIL PROTECTED]
wrote:
Hi all.
I have written a guestbook (I know, there's a ton out there), which until
recently did not get any spam, so now I'm trying to remove the spam-ad by
using preg_replace:
$melding = preg_replace('/(a href=\http:\/\/[^^]+)/i', This message is
temporarily cut-off due to spam-suspicion., $melding);
However, this only makes the problem half-solved, since I also want any text
before the link to be replaced with a message as stated above. The above
line removes anything after 'a href=//'.
Is there any way to make the above line to include anything before the a
href=// ?
I have tried different options, and yes, I have read the php manual on
preg_replace, but I might not have properly identified how to get the text
in front modified.
Thankful for any hints, tips, links, anything that helps :-)
Regards,
Martin Bach Nielsen
If you only want to replace the line including url(s) by a message, a
simple regular expression is enough.
?php
$melding = This is a spam.\n
. This is a a href=\http://www.example.com/\;example/a line.\n
. Another line.\n
. a href=\http://www.e.com/ads/show?n=123\;123/a (456)\n
. End Line.\n;
$melding = preg_replace(/.*?a href=\http:\/\/.+/i, This line is
temporarily cut-off due to spam-suspicion., $melding);
echo $melding;
?
The script above will output:
This is a spam.
This line is temporarily cut-off due to spam-suspicion.
Another line.
This line is temporarily cut-off due to spam-suspicion.
End Line.
--
Sorry for my poor English.
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