[SOLVED] Function not working

[master82]

In a form, when a user clicks the submit button the function below runs, however I get the following error:


[quote]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 gangwars where warID=[/quote]


[code]<?php
//...

case "viewsurrenders":
gang_staff_viewsurrenders();
break;

//...

function gang_staff_viewsurrenders()
{
global $db,$ir,$c,$userid,$gangdata;
if(!isset($_POST['subm']))
{
print "<form action='yourgang.php?action=staff&act2=viewsurrenders' method='post'>
Choose who to accept the surrender from.<br />
<input type='hidden' name='subm' value='submit' />
Faction: <select name='sur' type='dropdown'>";
$wq=$db->query("SELECT s.*,w.* FROM surrenders s LEFT JOIN gangwars w ON s.surWAR=w.warID WHERE surTO={$ir['gang']}");

while($r=$db->fetch_row($wq))
{
if($gangdata['gangID'] == $r['warDECLARER']) { $w="You";$f="warDECLARED"; } else { $w="Them";$f="warDECLARER"; }
$ggq=$db->query("SELECT * FROM gangs WHERE gangID=".$r[$f]);
$them=$db->fetch_row($ggq);
print "<option value='{$r['surID']}'>War vs. {$them['gangNAME']} (Msg: {$r['surMSG']})</option>";
}
print "</select><br /><input type='submit' value='Accept Surrender' /></form>";
}
else
{
$_POST['sur'] = abs((int) $_POST['sur']);
$q=$db->query("SELECT surWAR FROM surrenders WHERE surID={$_POST['sur']}");
list($_POST['war']) = $db->fetch_row($q);
$wq=$db->query("SELECT * FROM gangwars where warID={$_POST['war']}");
$r=$db->fetch_row($wq);
if($gangdata['gangID'] == $r['warDECLARER']) { $w="You";$f="warDECLARED"; } else { $w="Them";$f="warDECLARER"; }
$db->query("DELETE FROM surrenders WHERE surID={$_POST['sur']}");
$db->query("DELETE FROM gangwars WHERE warID={$_POST['war']}");
$ggq=$db->query("SELECT * FROM gangs WHERE gangID=".$r[$f]);
$them=$db->fetch_row($ggq);
$event=str_replace("'","''","<a href='gangs.php?action=view&ID={$ir['gang']}'>{$gangdata['gangNAME']}</a> have accepted the surrender from <a href='gangs.php?action=view&ID={$them['gangID']}'>{$them['gangNAME']}</a>, the war is over!");
$db->query("INSERT INTO gangevents VALUES('',{$ir['gang']},unix_timestamp(),'$event') , ('',".$r[$f].",unix_timestamp(),'$event')");
print "You have accepted surrender, the war is over.";
}
}

//...
?>[/code]

Any idea whats wrong? I suspect from the error output its something to do with a missing variable (warID)

Also I have printed/echo all the lines - and it seems there is no value for the lines after:

[code]list($_POST['war']) = $db->fetch_row($q);
[/code]

Any help would be welcome

[mansuang]

Try
[code]
$wq=$db->query("SELECT * FROM gangwars where warID='{$_POST[war]}'");
[/code]

[master82]

No sorry - I think the problem is that $_POST['war'] is not being assigned a value - could be wrong

[trq]

Please format your code in a way which is readable. Most programmers indent code in logical blocks, yours is just a blob of text.

[Jessica]

If you think the $_POST['war'] is empty, then check your form! when you process, do
print_r($_POST) and you will see all posted vars.

[marcus]

Woah, Jesi, never thought I'd see you here.

[master82]

solved - slightly edited the code

[Jessica]

Okay...