Use this:
$result = mysql_query("SELECT PASSWORD(" . $_POST['password'] . ")");
$password = mysql_result($result,0);
or just use mysql_fetch_row() or AS in your query so you don't have to
recreate that complex column name.
---John Holmes...
> -----Original Message-----
> From: Murat Ö. [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, September 22, 2002 9:33 AM
> To: [EMAIL PROTECTED]
> Subject: [PHP] mysql password function
>
> hi,
> i want to encode a string that users enter with mysql password
function.
> but
> sometimes this code works sometimes don't. mysql warns me:
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL
> result
> resource in ........
>
> the code is:
>
> $result=mysql_query("select password(".$_POST['password'].")");
> while ($p = mysql_fetch_array($result, MYSQL_ASSOC)):
> $pswrd=$p['password('.$_POST['password'].')'];
> endwhile;
>
> thanks...
>
>
>
>
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