Member Approval v 0.1

[ron7bg 0]

As I do not know much about PHP, I am wondering if somebody who knows more can help.

 

Is it a question of adding :

where member_level = '1'

somewhere in customers.php?

 

in admin/customers.php find (around line 1425) :

<?php

$search = '';

if (isset($HTTP_GET_VARS['search']) && tep_not_null($HTTP_GET_VARS['search'])) {

  $keywords = tep_db_input(tep_db_prepare_input($HTTP_GET_VARS['search']));

  $search = "where c.customers_lastname like '%" . $keywords . "%' or c.customers_firstname like '%" . $keywords . "%' or c.customers_email_address like '%" . $keywords . "%' and c.member_level = '1'";

}

 

and right after it add:

 

	if ($search =='') $search = "where member_level = '1'";

[osthinker 0]

i updated from 1.5 to 1.7

when i try to approve a client, i get this message in the admin

 

1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

select customers_language_id from customers where customers_id = 

[TEP STOP]

 

anyone?

 

thanks in advance

Edited July 23, 2008 by osthinker

[osthinker 0]

i updated from 1.5 to 1.7

when i try to approve a client, i get this message in the admin

 

1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

select customers_language_id from customers where customers_id = 

[TEP STOP]

 

anyone?

 

thanks in advance

 

 

found it in the german forum! thanks guys even that i did not understand a word from you ;)

 

in admin/members.php on line 163 find

 

$customer_language_id_qry = tep_db_query('select customers_language_id from ' . TABLE_CUSTOMERS . ' where customers_id = ' . $cID);

 

and replace it

 

$customer_language_id_qry = tep_db_query('select customers_language_id from ' . TABLE_CUSTOMERS . ' where customers_id = ' . $HTTP_GET_VARS['cID']);