I'm trying to run a script which uses variables passed via a form, like
this:
if($lcat_name)
{
$sql_query = "INSERT into category VALUES ('','$lcat_name')";
$result = mysql_query($sql_query);
echo "You add next category $lcat_name";
}
but the variable $lcat_name is empty unless I do this:
$lcat_n
No it's not.
It's a script called LinksCaffe
<http://www.hotscripts.com/cgi-bin/dload.cgi?ID=15062>.
In file catadd.php
"Leif K-Brooks" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Is it inside of a function?
>
> samug wrote:
>
> &
There's no php.sql, so I'll put it here.
I'm trying to do a search from mysql database like this:
if (!$inquiry = mysql_query("select id,link,heading,desc from links where
heading like '%$entry%' or desc like '%$entry%'
or keyword like '%$entry%'",$connection)){
print "Search was unsuccesful!
Could someone tell me why isn't this working?
$text = "And what did I tell you?";
list($one,$two,$three,$rest) = preg_split("/\s+/", $text);
print($rest);
The result would be "I" and not "I tell you?"
Why?
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Thanks.
But what if you don't know how many elements there will be ahead?
"Samug" <[EMAIL PROTECTED]> kirjoitti viestissä
news:20021109165530.58233.qmail@;pb1.pair.com...
> Could someone tell me why isn't this working?
>
> $text = "And what did I t
Hi,
I'm quite new to php and I'm trying to execute a file with
exec("executable.exe"); and nothing happens.
I have win2k with apache 1.3.23 and php 4.1.1 (ISAPI).
I read somewhere that running external programs with isapi is impossible. Is
that true? I can still run system commands, i.e. system("d
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