Using variable in variable name

I am writing a script to backup tables in a specified database. I first execute a query to get a listing of table names in that db(show tables). Then I loop through the results one row at a time, using exec to pass the name of the table to mysqldump. The problem is that the database name is defined in a variable at the top of the script, like so: $db = "DB"; The variable $db is then used in the query, and the subsequent exec call $show_q = "SHOW TABLES FROM $db"; $show_a = mysql_fetch_array($show_r); $show_rows = mysql_numrows($show_r); for ($i = 0; $i < $show_rows; $i++) { $show_a = mysql_fetch_array($show_r); exec("mysqldump -uroot --opt $db $show_a[Tables_in_$db] > ..\admin\dbbackups\\$show_a[Tables_in_$db].sql"); } But I am getting a parse error for Tables_in_$db - how do I correctly use the variable $db within the variable $show_a[Tables_in_$db]? Thanks, -Lisi