What is the error? there i something i always do when i have this problem,
try this:
error_reporting(E_ALL);
$sql = "INSERT INTO log SET term='$search', returns='$arrayword',
time=CURTIME(), date=CURDATE(), ip='$ip'");
echo $sql;
$logit = mysql_query($sql);
This makes all kind of notice about uninitialized variables and other things
that may cause problems, great for debugging, and the echo $sql prints the
sql that is sent to the client, here you can see if one of the variable yo
are sendding to the query have some kind of problem, like a ' in the middle,
which could cause problems in your script.
If you can't see the problem after this post the output of this code.
----- Original Message -----
From: "Axel IS Main" <[EMAIL PROTECTED]>
To: "PHP-DB" <[EMAIL PROTECTED]>
Sent: Thursday, February 26, 2004 9:05 AM
Subject: [PHP-DB] What's wrong with this query?
> I've just tried to do something I've done a thousand times. It does not
> work. I've checked all of the syntax, made sure the field and variable
> names are correct. No matter what I do it just doesn't work. The table
> remains empty. Here's the query:
>
> $logit = mysql_query("INSERT INTO log SET term='$search',
> returns='$arrayword', time=CURTIME(), date=CURDATE(), ip='$ip'");
>
> Now that doesn't look too difficult does it? Well, apparently it's
> impossible! I'm really hoping someone out there can see something that I
> missed.
>
> Nick
>
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