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

Your Answer:

Login to answer

Other forums

Gaining access to Drop Down Menu Variable
code is below, I am creating a drop down menu. I want to echo the value selected in the drop down me

Access website from only 1 computer...?
One of my customers wants his website to only be accessed by people in which they bought the website

Multi Dimensional Array Append
Hi Guys

I have a function that returns a multidimentional array eg

$result = functio

Parse error: syntax error, unexpected $end in
hello,
i receive the error :Parse error: syntax error, unexpected $end in C:xamppxampphtdocsh

tag inside php...
hi


Code: <?php
echo "<td width="$columnWidth%" align=&

email form (cannot find the problem)
Hello there. I've been having trouble with an email form. Can't find the problem really. I've tested

Recognising Revenue daily
Hello.
Has anyone come across daily recognition of deferred income?
We want to implement R

PRotect my POST DATA... FROM HTML ATTACKS
I need to like let's say

If a user submit's his Prayer and he put's <meta refresh stuf

Shuffle between users ??
I have multiple $users in table. I need to send them $message.

I need to send the next messag

question about header() security
is is safe to just use the header() function to redirect someone if they are, say, not logged in? or