Error when using a variable to select database table

[doc1355]

Hi,
I have a script that I need to use a variable to select the database table that I want to update or to get the results from. But for some reason when I use the variable I get error. If I replace the variable with the table name, it works fine.

Here is the part of the code related to this issue:
[CODE=php]<?php
$memberID = $_GET['cm'];
if ($memberID < 50001)
$data_table = 'new_members' &&
$auth_token = "first_toke n";
else
$data_table = 'old_members' &&
$auth_token = "second_tok en";



$find_customer = mysql_query("SE LECT * FROM '$data_table' WHERE id = $memberID");
while($customer _info = mysql_fetch_arr ay($find_custom er))
{
$customer_full_ name = $customer_info['first_name'] . " " . $customer_info['last_name'];
$customer_email _address = $customer_info['email'];
}
?>
[/CODE]

[pbmods]

Heya, Doc.

Try using backticks instead of quotes:
[code=mysql]
SELECT * FROM `$data_table` WHERE id = $memberID
[/code]

[doc1355]

I did that but same problem, still not connecting to the table.
I also tried with out quotes, but no luck.

[pbmods]

Heya, Doc.

What happens when you echo mysql_error()?

[doc1355]

I get multiple Undefined variable error messages.


Notice: Undefined variable: customer_email_ address in F:\USERS\userna me\www\live\for ms\registration \success.php on line 177

Notice: Undefined variable: customer_full_n ame in F:\USERS\userna me\www\live\for ms\registration \success.php on line 173

[code green]

Examine this a bit closer
[PHP]if ($memberID < 50001)
$data_table = 'new_members' &&
$auth_token = "first_toke n";
else
$data_table = 'old_members' &&
$auth_token = "second_tok en"; [/PHP]