Cannot Display Array from Select Statement + Login question

Posted on 16th Feb 2014 07:03 pm by admin

Hi,
I'm new to PHP but so far so goog. I was assigend a project and I'm very close to completion.

I have a site that logs you in, sets a cookie, and then what I would like to do is depending on the user loging in, certain records from a table are selected and displayed

I have the following code:

Code: // Connects to your Database
mysql_connect("localhost", "root", "escrma") or die(mysql_error());
mysql_select_db("rma_portal") or die(mysql_error());

//Set Login Variable with the UserID
$login = $_COOKIE['ID_my_site'];

echo "This is it set to $login

";

// Collects data from "users" table
$data = mysql_query("SELECT *
FROM users, cust_info
WHERE cust_info.Cust_ID = '.$login'")
or die(mysql_error());

// puts the "users" info into the $info array
$info = mysql_fetch_assoc($data);

//Test Print of Login ID passed value
echo $data; WORKS Resource id #3
echo $info["Cust_Name"]; DOES NOT WORK
echo "$Cust_Name"; DOES NOT WORK

echo "$login

"; WORKS
echo $info["Cust_Company"]; DOES NOT WORK


1 Question is why I can echo the $login but then I canot pull anything from the query?
Table users
Cust_ID, Username, Password
Table cust_info
Cust_ID, Cust_Name, Cust_Company, Cust_Phone.....

2 I don't know if using the Cookie variable to use the login is a good idea. What would be the best way to pass the username to next page so I can use it in the SELECT ststement. Should I use a _SESSION global?

Thanks in advance

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Code: [Select]<?php
//include shared codes

include '../lib/common.php';
include