Re: Using variable in variable name

Use the string-operator . if i understood your code right: exec('mysqldump -uroot --opt $db $show_a[Tables_in_$db] > ..\admin\dbbackups\' . $show_a[ 'Tables_in_' . $db ] . '.sql'); but using the exec-method seems very dangerous to me... .ma Lisi <lists@shemeshdirectory.co.il> wrote@25.05.2003 12:45 Uhr: > 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 >