tomer.
The code above is passed to another form for processing and is were the
insert data occurs. Using this method I can echo the form data easily
Hope this helps...
Aleks
-----Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Tuesday, December 23, 2003 4:33 AM
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Tuesday, December 23, 2003 4:33 AM
To: David Robley
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Re: Display Mysql Result in drop down list
Hi:
then now i'm trying to select a value from the drop down list, echo
Hi:
then now i'm trying to select a value from the drop down list, echo the value
i have selected and lastly INSERT the value into DB with the following :
Below's a snippet of code:
//retrieve all "tutor_name" from DB and display in drop down list
" .$row
["tutor_name"]. "";
}
$result = $db
In article <[EMAIL PROTECTED]>,
[EMAIL PROTECTED] says...
>
> Hi all,
>
> Right now I'm trying to retrieve one of the column "tutor_name" and display al
> the tutor's name in a drop down list. The problem now is, the drop down list
> only manage to display 1 record from that row instead of al
