Quotes missing around $value. Try instead:
mysql_query( "UPDATE tablename SET {$name} LIKE '{$value}' WHERE ID={$id}")
Of course, even though the code does not show, you have properly validated
user input before firing this query...<blink/>
HTH
Ignatius
_________________________
----- Original Message -----
From: "Robbie Staufer" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, September 24, 2003 5:20 PM
Subject: [PHP-DB] update db with variables
> Hi,
>
> I'm writing an application that will display the contents of a db in the
> browser, including the db-assigned id number. The user can then enter
> the id number in a form and the browser will display the record
> corresponding to that id number, with the data in editable text fields.
>
> The user edits the text fields and submits to an 'update' script using
> POST, and the update script puts the new values back into the database.
> In the update script, I'm trying to access the field names and values
> in the $_POST array, and update the db using a loop:
>
> while ($data = each($_POST)) {
> $name = $data[ 'key' ];
> $value = $data[ 'value' ];
> echo $name, $value;
> $result =
> mysql_query("UPDATE tablename SET $name = $value WHERE ID=$id") or die
> ("unable to update record");
> }
>
> The echo statement shows me that the script dies on the first time
> through the loop.
>
> The application must be on line this week. Does anyone have any
> suggestions?
>
> Many thanks,
> Robbie
>
>
>
> --
> -*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*
> Robbie Staufer
> NCAR/SCD
> 1850 Table Mesa Dr. Rm. 42
> Boulder, CO. 80305
> (303) 497-1836
>
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