> but using the exec-method seems very dangerous to me...
I'll second that.
Edward Dudlik
Becoming Digital
www.becomingdigital.com
----- Original Message -----
From: "heilo" <grillen@abendstille.at>
To: "PHP-DB" <php-db@lists.php.net>
Sent: Sunday, 25 May, 2003 06:24
Subject: 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
>
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