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Hi all, I have this piece of code which im using for my website but for a werid reason its saying the table that it selects the information out of doesnt exist when it does.

 

This is my code:

 

$locations = array('England','Japan','Colombia','Usa','South Africa','Mexico');
$i =0;
$count = count($locations);
while ($i < $count){
if ($i == "2"){ echo "<tr>"; }
$airport = mysql_fetch_object(mysql_query("SELECT * FROM airport WHERE location='$locations[$i]'"));
$ammo_hut = mysql_fetch_object(mysql_query("SELECT * FROM bf WHERE location='$locations[$i]'"));
$bank = mysql_fetch_object(mysql_query("SELECT * FROM bank WHERE location='$locations[$i]'"));
$people_in = mysql_num_rows(mysql_query("SELECT * FROM users WHERE location='$locations[$i]'"));
$dealer=mysql_fetch_object(mysql_query("SELECT * FROM dealership WHERE location='$locations[$i]'"));
$rest=mysql_fetch_object(mysql_query("SELECT * FROM rest WHERE location='$locations[$i]'"));
$casino_slots = mysql_fetch_object(mysql_query("SELECT * FROM casino WHERE location='$locations[$i]' AND type='Slots'")) or die (mysql_error()); // Line 168
$rest=mysql_fetch_object(mysql_query("SELECT * FROM rest WHERE location='$locations[$i]'"));
$casino_rps = mysql_fetch_object(mysql_query("SELECT * FROM casino WHERE location='$locations[$i]' AND type='RPS'")) or die (mysql_error());
$casino_race = mysql_fetch_object(mysql_query("SELECT * FROM casino WHERE location='$locations[$i]' AND type='Race'")) or die (mysql_error());
$bar = mysql_fetch_object(mysql_query("SELECT * FROM bar WHERE location='$locations[$i]'"));
$shop = mysql_fetch_object(mysql_query("SELECT * FROM shop WHERE location='$locations[$i]'"))  or die (mysql_error());

 

The error im getting is:

 

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/www/mafia-drift.co.cc/locations.php on line 168

Table 'deabot96_users.casino' doesn't exist

 

 

This isnt the full script, if needing to see it all ask :).

Thanks for your help.

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https://forums.phpfreaks.com/topic/201113-error-in-code/
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You need to investigate what your actual table name is v.s. what is in your query and attempt to find what is different between the two.

 

You either have a spelling error, a capitalization difference (assuming you are on an operating system that is case-sensitive), or you have some non-printing or special characters as part of the table name.

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https://forums.phpfreaks.com/topic/201113-error-in-code/#findComment-1055117
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