You've got too many parameters in the mysql_connect() and mysql_select_db()
functions. You don't need the %s stuff. Here's what it should be:
$conn = mysql_connect($connectie,$uid,$pwd);
mysql_select_db($database,$conn);
Good luck,
Tyler Longren
Captain Jack Communications
[EMAIL PROTECTED]
www.captainjack.com
----- Original Message -----
From: "Tommy Straetemans" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, November 29, 2001 8:52 AM
Subject: [PHP] database conection (newby)
> Hi
>
> I'am totaly new in php i work always with asp.
> I give the folowing hidden fields to a php script
> <input type=hidden name=connectie value="mysqlhost.mijndomein.be:3306">
> <input type=hidden name=uid value="U0498526">
> <input type=hidden name=pwd value="iwinfo59">
> <input type=hidden name=database value="D0498526">
>
> and in my php script i try to make my connection as follows:
>
> $conn = mysql_connect('%s','%s','%s',$connectie,$uid,$pwd);
> mysql_select_db('%s',$database,$conn);
>
> but i keep getting this error:
> Warning: Wrong parameter count for mysql_connect() in
> /home/users/mijndomein.be/admin/x.php on line 27
>
> What do i wrong or is there a method to make my connection with a include
> file?
>
>
>
>
> Tommy Straetemans
>
>
>
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