Hi i have a question. I got this query :

$hostid_query = "SELECT hostid FROM webhosts ORDER BY hostid DESC";

now then, i following that with this :

$result = mysql_query($hostid_query);
$data = mysql_fetch_array($result);
$last_hostid = $data['hostid'];

Now then, i thought that it would put the highest hostid into $last_hostid by it doesn't

the last host_id is 22, and it puts 7 in there for some reason

anyone know what im doing wrong

thanx

Couple of things to look at:

1) What is the data type of the hostid field?

2) What result do you get if you do this query from the mysql client or from phpmyadmin?

Richard

Try running the query in a program such as phpMyAdmin and that a look at all the results returned. The problem may be more visable then.

And if you don't like the web-based-idness of phpMyAdmin or don't have access, you can use something like MySQL-Front - a windows based program, really nice work. Download mirror and forum/informational links are available at .mysqlfront.referable.com

I use MySQL Front and love it. That's how I trouble shoot queries.

It looks like it should work but then you don't see a mis-spelling or a wrong command or what ever.

Hrm, I believe it should be


$q = "SELECT hostid FROM webhosts ORDER BY hostid DESC LIMIT 1";

$result = mysql_query($q);
$data = mysql_fetch_array($result);
$last_hostid = $data['hostid'];

found the problem

i set hostid as VARCHAR not tinyint

Thanx why it was displaying the data wrong

thanx for the comments and sugestions

thanx :)