Using variable in variable name

[Lisi <lists@shemeshdirectory.co.il>]

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


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