"')";
>> $result = mysql_query($query);
>
> That one did the trick!
>
> Thanks James and Wouter for helping me out!
>
> Frank
> ----- Original Message -----
> From: "esctoday.com | Wouter van Vliet" <[EMAIL PROTECTED]>
> To: "Fr
ot;;
> $result = mysql_query($query);
>
> # and what else you'd want to do ...
> };
>
> would just make even a little more sense .. some discussion could be about
> using $_POST['test1'][$i] to get the value again, or just using the $Value
> variable created by t
using the $Value
variable created by the foreach calls...
-Oorspronkelijk bericht-
Van: Frank Keessen [mailto:[EMAIL PROTECTED]
Verzonden: zondag 15 juni 2003 21:44
Aan: Frank Keessen; James Hicks
CC: [EMAIL PROTECTED]
Onderwerp: Re: [PHP] Capturing the input of dynamic form fields (part
II)
nk
- Original Message -
From: "Frank Keessen" <[EMAIL PROTECTED]>
To: "James Hicks" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Sunday, June 15, 2003 8:59 PM
Subject: Re: [PHP] Capturing the input of dynamic form fields (part II)
> Hi Ja
uot; <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Sunday, June 15, 2003 4:14 PM
Subject: Re: [PHP] Capturing the input of dynamic form fields (part II)
> Whenever I am having problems with SQL queries I always display the query
on
> the page with all the replaced variables
Whenever I am having problems with SQL queries I always display the query on
the page with all the replaced variables so I can make sure that the query
makes sense. If it passes the MySQL parser in my brain and still doesn't
work, I will copy the displayed query into the command line mysql monit
Dear all,
Refering to my first e-mail with the same subject; I've reworked the code to this:
But the only problem is; Nothing is written into the mysql database! And there are no
error displayed..
Can someone have a look at my code?
";
echo "# $i";
}
?>
Thanks very much,
Frank
Dear all,
Can you help me with the following or point me to the right direction;
I'm receiving value's of two fields called test1 and test2. Both fields can have
multivalue's. The first field called test1 is going o.k. with the following code;
if(is_array($_POST['test1'])) {
foreach($_POST['tes
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