[SOLVED] Return error

[Minase]

$ranking = mysql_query("SELECT * FROM `" . DBPREFIX . "users` ORDER BY Capture DESC Limit 0,50") or die(mysql_error());
while($r = mysql_fetch_array($ranking)){
echo '
<tr>
<td align="center">'.$r['Capture'].'</td>
</tr>
';
}

 

can someone tell me what is wrong with my code please?

 

look at my image :

http://img84.imageshack.us/img84/2832/unbenanntdt0.png

 

booty = Capture column

 

it doesnt return normal results

thanks

[trq]

it doesnt return normal results

 

That means nothing to us. Define normal.

[Minase]

look at picture...

first return = 950

second = 50

third = 439

 

normal it should be

1 = 950

2 = 439

3 = 50

 

cause it is ordered by Capture descending...

[JasonLewis]

Uh, shouldn't you be getting all the related data to each user and echoing it in the one loop?

[Minase]

i need every value there

anyway topic solved.

i found mysql treat some columns type as varchar

with int it work just great

in past from what i remember even varchar worked fine.. but oh well maybe is just me  ???

thanks all