php mysql, variables for tablenames in querys

d0m1n1c

ok, so i have this:

list($table2, $uid) = explode("-", $referer)
$addreferal = mysql_query("UPDATE `artists` SET referrals = referrals + 1 WHERE id = '$uid'",$db);

where the table is artists, how would i set it to be the value of $table2.

Ive tried everything, it works fine when its used in the where clause but i cant seem to get it working in the table name. its not the variable, i know thats fine and assigned, because when i echo the variable before it is used in the query, it return what i want ok, but for some reason, it wont workk as a table name.

please help

thanks

dom

wilku

list($table2, $uid) = explode("-", $referer) ;
$addreferal = mysql_query("UPDATE `$table2` SET referrals = referrals + 1 WHERE id = '$uid'",$db);

There was a semicolon missing at the end of first line, but it's probably just a typo.

If it will not work, try

list($table2, $uid) = explode("-", $referer) ;
$sql = "UPDATE `$table2` SET referrals = referrals + 1 WHERE id = '$uid'";
echo $sql;
$addreferal = mysql_query($sql,$db);

and see if the query shown with echo, will work in mysql client or phpmyadmin. If id is of type integer, I wouldn't put quotes around $uid.

d0m1n1c

ok, still sort of along the same lines almost, but i didnt want to start a new thread.

sooooo im not sure if what you told me worked (not the semi colon, that was there but i didnt put it in the forum (dunno y)

but anyway, i tested my split command.

				
list($table2, $uid) = explode("-", $referer);
echo($table2);
echo("<P>next should be another variable</P>");
echo($uid);

and when i load it (when referer = a-1) i get:

\\"a

next should be another variable

1\\"

That leads me to believe that $table2 = \\"a and $uid = 1\\"

Do do i get rid of the \\" in it so then i can test if the table thing works?>

Thanks

Dom