[SOLVED] QUERY ERROR, but it seems fine?

[Lt_Wolf]

I'm making a Hall Of Fame for someone's game and im coming across this problem.  This is the error messege:

 

QUERY ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

Query was SELECT * FROM userstats WHERE userid=

 

just a note:  There are 'userid' rows for both the users and the userstats tables.

 

 

function hof_strengthg()

{

global $db,$ir,$c,$userid, $myf;

print "Showing the 20 users with the highest strength gain this month<br />

<table width=75% border=1 class='tablejf1'><tr background='http://suburban-assault.net/header.jpg'> <th>Pos</th> <th>User</th><th>Amount</th> </tr>";

$q=$db->query("SELECT u.*,g.* FROM users u LEFT JOIN userstats us ON u.userid=us.userid LEFT JOIN gangs g ON g.gangID=u.gang WHERE u.user_level != 0 AND u.user_level != 2 $myf ORDER BY us.strGAIN DESC,u.userid ASC LIMIT 20");

$gh=$db->query("SELECT * FROM userstats WHERE userid={$r['userid']}");

$p=0;

while($r=$db->fetch_row($q))

{

$p++;

if($r['userid'] == $ir['userid']) { $t="<b>";$et="</b>"; } else { $t="";$et=""; }

print "<tr> <td>$t$p$et</td> <td>$t{$r['gangPREF']} {$r['username']} [{$r['userid']}]$et</td><td>{$gh['strGAIN']}</td> </tr>";

}

print "</table>";

}

[trq]

I don't see anywhere in the function where you define the array $r.

[Lt_Wolf]

an obvious mistake, but the error is still showing.  New code:

 

($r is declared in a file included for the Hall Of Fame page)

 

function hof_strengthg()

{

global $db,$ir,$r,$c,$userid, $myf;

print "Showing the 20 users with the highest strength gain this month<br />

<table width=75% border=1 class='tablejf1'><tr background='http://suburban-assault.net/header.jpg'> <th>Pos</th> <th>User</th><th>Amount</th> </tr>";

$q=$db->query("SELECT u.*,g.* FROM users u LEFT JOIN userstats us ON u.userid=us.userid LEFT JOIN gangs g ON g.gangID=u.gang WHERE u.user_level != 0 AND u.user_level != 2 $myf ORDER BY us.strGAIN DESC,u.userid ASC LIMIT 20");

$gh=$db->query("SELECT * FROM userstats WHERE userid={$r['userid']}");

$p=0;

while($r=$db->fetch_row($q))

{

$p++;

if($r['userid'] == $ir['userid']) { $t="<b>";$et="</b>"; } else { $t="";$et=""; }

print "<tr> <td>$t$p$et</td> <td>$t{$r['gangPREF']} {$r['username']} [{$r['userid']}]$et</td><td>{$gh['strGAIN']}</td> </tr>";

}

print "</table>";

}

[Lt_Wolf]

anyone?

[GingerRobot]

Well, presumably you are getting a different error now? Since r is defined, the user id is presumably part of the query now?

[Lt_Wolf]

the error is the same

 

"QUERY ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

Query was SELECT * FROM userstats WHERE userid="

[DyslexicDog]

What do you get when you echo $r['userid']?

 

and I'm pretty new to mysql statements but why are you using curly braces in your select statement?

 

could this work?

$gh=$db->query("SELECT * FROM userstats WHERE userid=$r['userid']");

[Psycho]

What does this output to the page:

 

<?php

echo "<pre>";
print_r($r);
echo "</pre>";

?>

[teng84]

the error is the same

 

"QUERY ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

Query was SELECT * FROM userstats WHERE userid="

obviously empty variable or something like that

SELECT * FROM userstats WHERE userid=" <=--- no value in where condition  check your variable or print your query

[Lt_Wolf]

thanks for the help everyone, I have found a solution myself