SQL Syntax Error!

[Bradley99]

Hi all, This error is in the Inboxread.php file at the part where players accept an invite to join an Organised Crime....

Original Error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/stayway1/public_html/www.one-mafia.com/eng/inboxread.php on line 99

Here's the code lines 90-110 (line 99 in bold tags):

if (strip_tags($_POST['Accept_OC'])){
$oc_id=strip_tags($_POST['oc_id']);
if (strip_tags($_POST['place']) == "we"){
$use="we_inv";
$a="we";
$query= "SELECT * FROM oc WHERE we_inv='$username' AND id='$oc_id'";
}elseif (strip_tags($_POST['place']) == "ee"){
$use="ee_inv";
$a="ee";
$query= "SELECT * FROM oc WHERE ee_inv='$username' AND id='$oc_id'";

}elseif (strip_tags($_POST['place']) == "driver"){
$use="driver_inv";
$a="driver";
$query= "SELECT * FROM oc WHERE driver_inv='$username' AND id='$oc_id'";
}

$round=mysql_query($query);
[b]$check=mysql_num_rows($round);[/b]
if ($check != "0"){
echo "You are now in that OC";
mysql_query("UPDATE `oc` SET `$a`='$username' WHERE `id`='$oc_id'");
mysql_query("UPDATE `users` SET `oc`='1' WHERE `username`='$username'");
}
}
if (strip_tags($_POST['Yes_street'])){
$race_id=strip_tags($_POST['race_id']);
if ($fetch->street != "0"){ echo "Your in a race."; }elseif ($fetch->street == "0"){

if ($fetch->last_race >= time()){ echo "You cant do anouther race yet."; }elseif ($fetch->last_race < time()){

And here's the 'Actual' error:

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 'Resource id #5' at line 1

Any help is appreciated.

[dagon]

this query failed

$query= "SELECT * FROM oc WHERE driver_inv='$username' AND id='$oc_id'";

debug it

$round = mysql_query($query) or die("<b>A fatal MySQL error occured</b>.\n<br />Query: " . $query . "<br />\nError: (" . mysql_errno() . ") " . mysql_error());

[Bradley99]

Yeah i already did and got the last error i posted, here it is again...

A fatal MySQL error occured.
Query: Resource id #5
Error: (1064) 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 'Resource id #5' at line 1

[dagon]

ah well see the query? does that make sense to you?

[Bradley99]

Yeah, Select from oc table where the driver/we/ee is the user and the id is the oc id..

Here's the OC code from where the message is sent...

if ($oc->we != "0"){
echo "You need to kick the Weapons expert before you invite someone else.";
}elseif ($oc->we == "0"){


$invite_text="<form name=invite_to_oc method=post action=?accept=oc&oc_id=$oc->id>
  <div align=center>You have been invited to join $username s Getaway in $lead->location, please 
    choose one of the following options:<br>

<input name=Decline class=button type=submit id=Decline value=Decline>
| 
<input name=Accept_OC type=submit class=button id=Accept value=Accept>
  </div>
</form>"; 

mysql_query("INSERT INTO `inbox` ( `id` , `to` , `from` , `message` , `date` , `read` )
VALUES (
'', '$inv_username', '$inv_username', '$invite_text', '$date', '0'
)");
echo "Weapons expert invited";
mysql_query("UPDATE oc SET we_inv='$inv_username' WHERE id='$oc->id'");

[dagon]

currently $query="Resource id #5"

when you get to here:

$round=mysql_query($query);

that's the result of another query which is not in the code you have posted.

[Bradley99]

Which code? The first one i posted or second?

[dagon]

the first