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Author	Thread
Ensellitis Paranoid (IV) Inmate From: Kansas City, MO , USA Insane since: Feb 2002	posted 03-24-2010 05:27 I have been fighting this for hours it seems... I have this function: code: function levelwith3up($id,$threeup1db,$threeup2db,$threeup3db) { if($threeup1db == 0 && $threeup2db == 0 && $threeup3db == 0) mysql_query("UPDATE members SET level = 1 WHERE ID = '$id'"); if($threeup1db == 1 && $threeup2db == 1 && $threeup3db == 1) mysql_query("UPDATE members SET level = 2 WHERE ID = '$id'"); if($threeup1db == 2 && $threeup2db == 2 && $threeup3db == 2) mysql_query("UPDATE members SET level = 3 WHERE ID = '$id'"); if($threeup1db == 3 && $threeup2db == 3 && $threeup3db == 3) mysql_query("UPDATE members SET level = 4 WHERE ID = '$id'"); if($threeup1db == 4 && $threeup2db == 4 && $threeup3db == 4) mysql_query("UPDATE members SET level = 5 WHERE ID = '$id'"); } It has an odd problem... If the person is level 0 or 1, the update happens right away. However, if they hit level 3, I have to refresh the list to see the update. Am I missing something here? Am I using the right method for this? I'm at a loss...
Tyberius Prime Maniac (V) Mad Scientist with Finglongers From: Germany Insane since: Sep 2001	posted 03-24-2010 10:13 well, I don't see an obvious error, and I can't follow your 'see the update right away' description, but I'd start this differently and remove the duplication. code: function levelwith3up($id,$threeup1db,$threeup2db,$threeup3db) { if (($threeup1db == $threeup2db && $threeup1db == $threeup3db) && ($threeup1db >=0 && $threeup1db <= 4) { $new_val = intval($threeup1db) + 1; $id = mysql_real_escape_string($id); //seriously, does PHP still not have a template based sql engine???! mysql_query("UPDATE members SET level = $new_val WHERE ID = '$id'"); } else { error_occurred... } } ->Tyberius Prime
Petskull Maniac (V) Mad Scientist From: 127 Halcyon Road, Marenia, Atlantis Insane since: Aug 2000	posted 03-25-2010 08:05 Pardon my ignorance, but what is this code supposed to do?
revjx2 Obsessive-Compulsive (I) Inmate From: Insane since: Aug 2009	posted 03-26-2010 18:11 I was going to say about escaping $id or otherwise checking that it's not something nasty you're not expecting, beyond that I don't see why it would behave differently!
Ensellitis Paranoid (IV) Inmate From: Kansas City, MO , USA Insane since: Feb 2002	posted 03-29-2010 05:09 i think it was my debug script causing the problems. it works as it should now. thanks tyberious, you are still the man

Topic awaiting preservation: Stupid functions...\$

Author

Thread

Ensellitis
Paranoid (IV) Inmate

From: Kansas City, MO , USA
Insane since: Feb 2002

posted 03-24-2010 05:27

I have been fighting this for hours it seems...

I have this function:

code:

function levelwith3up($id,$threeup1db,$threeup2db,$threeup3db) {
	if($threeup1db == 0 && $threeup2db == 0 && $threeup3db == 0)
		mysql_query("UPDATE members SET level = 1 WHERE ID = '$id'");
	if($threeup1db == 1 && $threeup2db == 1 && $threeup3db == 1)
		mysql_query("UPDATE members SET level = 2 WHERE ID = '$id'");
	if($threeup1db == 2 && $threeup2db == 2 && $threeup3db == 2)
		mysql_query("UPDATE members SET level = 3 WHERE ID = '$id'");
	if($threeup1db == 3 && $threeup2db == 3 && $threeup3db == 3)
		mysql_query("UPDATE members SET level = 4 WHERE ID = '$id'");
	if($threeup1db == 4 && $threeup2db == 4 && $threeup3db == 4)
		mysql_query("UPDATE members SET level = 5 WHERE ID = '$id'");
}

It has an odd problem... If the person is level 0 or 1, the update happens right away. However, if they hit level 3, I have to refresh the list to see the update. Am I missing something here? Am I using the right method for this?

I'm at a loss...

Tyberius Prime
Maniac (V) Mad Scientist with Finglongers

From: Germany
Insane since: Sep 2001

posted 03-24-2010 10:13

well, I don't see an obvious error, and I can't follow your 'see the update right away' description,
but I'd start this differently and remove the duplication.

code:

function levelwith3up($id,$threeup1db,$threeup2db,$threeup3db) {
               if (($threeup1db ==  $threeup2db && $threeup1db == $threeup3db) &&
                    ($threeup1db >=0 && $threeup1db <= 4)
               {
                   $new_val = intval($threeup1db) + 1;
                   $id = mysql_real_escape_string($id); //seriously, does PHP still not have a template based sql engine???!
		mysql_query("UPDATE members SET level = $new_val WHERE ID = '$id'");
               }
              else
              {
              error_occurred...
            }
}

->Tyberius Prime

Petskull
Maniac (V) Mad Scientist

From: 127 Halcyon Road, Marenia, Atlantis
Insane since: Aug 2000

posted 03-25-2010 08:05

Pardon my ignorance, but what is this code supposed to do?

revjx2
Obsessive-Compulsive (I) Inmate

From:
Insane since: Aug 2009

posted 03-26-2010 18:11

I was going to say about escaping $id or otherwise checking that it's not something nasty you're not expecting, beyond that I don't see why it would behave differently!

Ensellitis
Paranoid (IV) Inmate

From: Kansas City, MO , USA
Insane since: Feb 2002

posted 03-29-2010 05:09

i think it was my debug script causing the problems. it works as it should now. thanks tyberious, you are still the man