Weird query result

JinF

I am trying to do a simple query, but instead of a numeric value, I get "Resource id #34".

Here is the code:

$newdodgemod1 = "select default_dodgemod from phpbb_classes where class='$newclass'";
$newdodgemod2 = $db->sql_query($newdodgemod1);

Any idea what is wrong with the code? All field exist and the variables are set correctly. I don't understand why I am getting this error.

Pieman86

You need to use mysql_fetch_object or something similar on $newdodgemod2:

http://www.php.net/mysql_fetch_object

JinF

Using

$newdodgemod3 = mysql_fetch_object($newdodgemod2);

I get the result of "Object".

And when I use

$newdodgemod3 = $db->sql_fetchrow($newdodgemod2);

I get "Array" as a result.

Did I use the wrong syntax?

Pieman86

No, that sounds right. 🙂

Have a look at the examples on php.net if you're not sure. fetch_object transforms the resource returned by your database into a PHP object. You should then be able to get the field you want like this:

$result4 = $newdodgemod3->default_dodgemod;

fetch_row should work as well. You can use:

$result4 = $newdodgemod3[0];

(you only selected one column from your database, so it'll be the 0th element in the array)

JinF

Thanks for the help so far!

Using

$newdodgemod2 = $db->sql_query($newdodgemod1);
$newdodgemod3 = mysql_fetch_object($newdodgemod2);
$newdodgemod4 = $newdodgemod3->default_dodgemod;

I still get the "Object" instead of the numeric value in the field.

Using

$newdodgemod2 = $db->sql_query($newdodgemod1);
$newdodgemod3 = $db->sql_fetchrow($newdodgemod2);
$newdodgemod4 = $newdodgemod3[0];

I get "Array" instead of the numeric value in the field.

I can't really see my error... :queasy:

bubblenut

What's the variable $db? Can you show us the entire code?

JinF

$db = new sql_db($dbhost, $dbuser, $dbpasswd, $dbname, false);

Basically my DB connection.

The whole code is almost 1000 lines and only this part is not working.

Pieman86

'tis odd. Can you do print_r() on the array you get after using fetch_row?