Raymond,
What sort of error are you getting?
One of the most useful tools in your toolbox is the print() function.
Try doing something like this:
$sSQL = "SELECT '$car' FROM varetabell WHERE carid='$carid'";
print ($sSQL);
$sql = mysql_query($sSQL);
$myrow = mysql_fetch_array($sql);
$x = $myr
t; <[EMAIL PROTECTED]>
To: "Raymond Lilleodegard" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Friday, January 25, 2002 8:11 PM
Subject: RE: [PHP-DB] Another dynamic sql.
> Seems like it will NOT work because you are selecting $car which is
already
> chosen by the user
urhan
-Original Message-
From: Raymond Lilleodegard [mailto:[EMAIL PROTECTED]]
Sent: Friday, January 25, 2002 1:48 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Another dynamic sql.
Hi all!
I have this form with some choices:
Enter how many cars you want:
ford
bmw
mercedes
And then I am try
Hi all!
I have this form with some choices:
Enter how many cars you want:
ford
bmw
mercedes
And then I am trying to get the price out of a table in my database with
this code:
$sql = mysql_query("SELECT '$car' FROM varetabell where carid='$carid' ");
$myrow= mysql_fetch_array($sql);
$x = $
