Resource id #153

Client Joined: Oct 07, 2003 Posts: 735 Location: Ohio

I am working on another page. I am pulling only one record from a table and am getting this Resource id #153 error.

I don't understand it. All I want to do is get the div_name that corrosponds to the div_commander who is logged in at the time.

Code:

$nukeusername = $userinfo['username'];  


  $sql = ("SELECT div_name FROM " . $prefix . "_eto_divisions WHERE div_commander ='$nukeusername'"); 


  $mydiv = $db->sql_query($sql);

When I echo $mydiv I get the Resource id error.

I ran across this in my research.

Quote:

When you get a result resource from mysql_query, you must then use one of the mysql functions that take mysql results resources as input (an argument) and processes them. A common one to use is mysql_fetch_assoc.

I still don't understand.

Posted: Mon Aug 07, 2006 2:16 pm

I believe mydiv is a "result identifier" that you can use to access a row in the database.

At any rate I think what you need is something like:

$row = $db->sql_fetchrow($mydiv);

Then I think you can retrieve the contents of the field with

$div_name = $row['div_name'];

you are just skipping a necessary step.

If you look in the /db directory at the mysql.php file you'll see that it does a fetch_array when you call for $db->sql_fetchrow. And we know from MYSQL documentation that the fetch_array returns an associative array and an associative array lets you get at values thru the field names.

MYSQL experts feel free to correct me if I've munged anything, I'm learning this as I go.

Posted: Mon Aug 07, 2006 3:22 pm

I'm no expert but the people at php.net seem to agree completely. Wink

Posted: Mon Aug 07, 2006 7:00 pm

I ended up doing

Code:




 $sql = "SELECT div_id FROM " . $prefix . "_eto_divisions 


  WHERE div_commander ='$nukeusername'"; 


  $result = $db->sql_query($sql);  


  $mydiv = $db->sql_fetchrow($result);


  $div_id = $mydiv[div_id];