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Posted by Bruce A. Julseth on 03/20/07 01:00
"Jeff" <jeff@quixion.net> wrote in message
news:1174344779.712020.258000@p15g2000hsd.googlegroups.com...
> Get rid of the @ in front of the $db = new mysqli( ...
>
> That is suppressing any errors on that line of code. Then it should
> show you in your browser what the error is and then we can help you
> further.
>
> Also, you can include $Database as the 4th parameter in your mysqli
> constructor to connect to that specific database.
>
Okay, I now have
$Host = "localhost";
$User = "Fred";
$Database = "house";
$Password = "mypw"
echo "before mysqli<br />Host: " . $Host . "<br />" . $User . "<br />" .
$Database;
$db = new mysqli($Host, $User, $Password);
echo "Connection is " . mysqli_connect_errno();
I didn't add the database to the mysqli parameter list.
I still never got to the 2nd echo statement. Firefox gives me a "done" in
the lower left corner. The "mysqli" is the very first MySQL command I
execute in my program. Do I need "Create" or "instantiate" something first?
I'm running PHP 5.2.0 (re: phpinfo()) and MySQL Server 5.0 [I'm not sure
what version. How can I find out?]
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