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$charinfo="SELECT name from `user_stats` WHERE User='$sname'";
$char=mysql_result(mysql_query($charinfo),0);

 

What this should do, and actually does in other parts of my code, is grab the name and then throw it into $char. Username and Char name are different.

The result is table does not exist. HOWEVER! The catch is, the output for the error message is "Table 'DB.$sname' doesn't exist"(DB is my actual database name and $sname is the actual called name, not the variable itself.) which is confusing the heck out of me. I haven't been able to find any reason why that would be happening. This is in MySQL 5.0, however I'm not sure if it's applicable or not. I'm sure this is just a foolish error on my part. However, any help would be greatly appreciated.

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don't use mysql_query inside of another function, I have it can cause problems (it has happend to me).

 

$charinfo="SELECT name from `user_stats` WHERE User='$sname'";
echo $charinfo; // Check what is being used as a query
$sql = mysql_query($charinfo)or die(mysql_error()); // See what error mysql gave us
$char=mysql_result($sql, 0);

 

edit:

Mysql tables are case sensitive.

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I've managed to track the problem down. As I said in my post, the snippet itself was used a couple other times. I'm using multiple tables and databases, so in calling a different database, I found my errors. At first, upon reviewing my code for the 1000th time, I noticed a missing piece of code calling the correct database. Thanks to all who attempted to solve my woes. This is why I should stop using my own code snippets. I forget about the small stuff. On a side note, when I was reviewing my code, again, I noticed the problem itself was actually below the chunk of code that I posted here. It was just that, in order to get to the erroneous code, the chunk had to be true. Anyway, thanks again for your aid.

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