My example was a litle confusing here I see.

My table have multiple price columns. something like this:

carid         convertible       stationvagon       etc. etc
1                100                100                    100



----- Original Message -----
From: "Gurhan Ozen" <[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  $car which is
already
> chosen by the user (i.e. which is a known value). I don't know how you
laid
> out your table but shouldn't your query be something like "SELECT price
from
> FROM varetabell where carid=$car"  ??
>
> Gurhan
>
> -----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:
>
> <form>
> Enter how many cars you want:<input type="text" name="number" >
> <select size="1" name="car">
> <option selected>ford</option>
> <option>bmw</option>
> <option>mercedes</option>
> </select>
>
> 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 = $myrow["$car"];
>
> $price = $x * $number;
>
>
> Shouldn't this work?  Or am I missing something here?
>
>
>
> Best regards Raymond
>
>
>
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