Re: Using variable in variable name

Lisi <lists@shemeshdirectory.co.il> · Sun, 25 May 2003 20:24:55 +0200

Why? Is there a better method?  There isn't really any user input, the 
script is set up by the site admin, and run via cron each day. I'm only 
using PHP because I'm not that great with Perl, although I have a feeling 
the script would be pretty similar.

-Lisi

At 11:03 AM 5/25/03 -0400, Becoming Digital wrote:
> 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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