Need Help with PHP code

[jazmine]

Hi,

I'm trying to write a script that allows for multiple users (managers from various stores) the ability to access their store information by typing in a identification number. All the information, from all the different stores, would be within 1 single database.

If a manager with an ID 7200 types in 7200 they will receive all the information from the database with the ID 7200.

I have written a script that allows a customer the ability to access his/her information from the database, but I'm stuck on how to allow someone the ability to access multiple customers information.

All the customers for a particular store would be listed under a number (e.g. 7200)
But how would you suggest I go about writing a script that allows the input of 7200 access to 200+ customers?

Do I need to have a table in MySQL for each individual store?

I'm quite lost on what to do...

Thanks for any help.

[tomhath]

Kind of hard to follow your question from what you posted.

Post the table definitions (table names and columns), a few rows of example data, and what you would like the output to be for the example data.

[jazmine]

Sorry about not being clear.

"Post the table definitions (table names and columns), a few rows of example data, and what you would like the output to be for the example data."

Here's what I would like to achieve.

The script I'm interested in writing would allow multiple users, who each have unique identification numbers, access to their customers within my database. In my database, I have a single table (called 'retailers') with a variety of fields (store_id, name, email, repair_number, etc.); the 'store_id' will be used to pull all the records from the database that has that particular number.

For instance, if Manager 1 has 7200 for his 'store_id', then when he types in 7200 in the 'login' page, he can access all 200 of his customers, each identified as 7200 in the database but each with varying information for each field. if Manager 2 has 8200 then he uses 8200 in the login page to access all 50 of his customers, etc.

This is the code I have written that allows for a single customer to access his/her account information:

if(isset($_POST['submit'])) {
	$jobnumber=$_POST['jobnumber'];

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("", $con);

$query = "SELECT * FROM customers WHERE job_number='$jobnumber' OR email='$jobnumber'";

$result = mysql_query($query);

My table is called: "retailers"
The fields are: "store_id, name, email, repair_number, in_date, est_date"
Some example rows:

"7200", "George Smith", [e-mail], 644305, 3/4/08, 5/6/08;
"7200", "Soxi Young", [e-mail], 684505, 3/4/08, 5/5/08;
"8200", "Mike Thomas", [e-mail], 644305, 3/4/08, 5/6/08;
"8200", "Robert Halt", [e-mail], 611110, 3/4/08, 5/5/08;

What I want the various managers to be able to see is exactly that. I just can't figure out what the code would be that would allow for 20 or so managers the ability to access their customers by entering their store_id.

Hopefully I did better that time.

Thanks again.

[HalfaBee]

You need to setup a form that the managers enter their store_id.

Then the sql looks like this

$store_id=mysql_real_escape_string( $_POST['store_id'] );
$query = "SELECT * FROM customers WHERE stor_id='$store_id' ";

[jazmine]

I tried this:

<?php


if(isset($_POST['submit'])) {

$store_id=mysql_real_escape_string( $_POST['store_id'] );

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("", $con);

$query = "SELECT * FROM retailers WHERE store_id='$store_id' "

$result = mysql_query($query);

echo "<table class='sample'>";
while ($row = mysql_fetch_array($result))  {

	//store id

	echo "<th>Store ID</th>";

	echo "<td>" . $row['store_id'] . "</td>";
	echo "</tr>";

	//name

	echo "<th>Customer Name</th>";

	echo "<td>" . $row['name'] ."</td>";
	echo "</tr>";

	//email

	echo "<th>Email</th>";

	echo "<td>".$row['email']."</td>";
	echo "</tr>";

	//repair number

	echo "<th>Repair Number</th>";

	echo "<td>".$row['repair_number']."</td>";
	echo "</tr>";

	//in date

	echo "<th>In Date</th>";

	echo "<td>".$row['in_date']."</td>";
	echo "</tr>";


	//est date

	echo "<th>Estimated Date</th>";

	echo "<td>".$row['est_date']."</td>";
	echo "</tr>";


}
	echo "</table>";
}
?>

But I'm getting a Parse Error on line 116: $result = mysql_query($query);

Don't know why that would be.

Thanks again.

[jaderobbins]

well one thing I see is that you aren't defining a database. Your table is also in a database that you need to define in your mysql_select_db statement

[jazmine]

I figured out the Parse error just now.

<?php

if(isset($_POST['submit'])) {

$store_id=mysql_real_escape_string( $_POST['store_id'] );

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("parsec", $con);

$query = "SELECT * FROM retailers WHERE store_id='$store_id' ";

$result = mysql_query($query);

echo "<table class='sample'>";
while ($row = mysql_fetch_array($result))  {

	//store id

	echo "<th>Store ID</th>";

	echo "<td>" . $row['store_id'] . "</td>";
	echo "</tr>";

	//name

	echo "<th>Customer Name</th>";

	echo "<td>" . $row['name'] ."</td>";
	echo "</tr>";

	//email

	echo "<th>Email</th>";

	echo "<td>".$row['email']."</td>";
	echo "</tr>";

	//repair number

	echo "<th>Repair Number</th>";

	echo "<td>".$row['repair_number']."</td>";
	echo "</tr>";

	//in date

	echo "<th>In Date</th>";

	echo "<td>".$row['in_date']."</td>";
	echo "</tr>";


	//est date

	echo "<th>Estimated Date</th>";

	echo "<td>".$row['est_date']."</td>";
	echo "</tr>";


}
	echo "</table>";
}
?>

I'm getting three errors:

Notice: Undefined index: storeid on line 5 = $store_id=mysql_real_escape_string( $_POST['store_id'] );

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'root'@'localhost' (using password: NO) in on line 5

Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in on line 5

I don't know. I'm befuddled.

[HalfaBee]

The first thing to do when befuddled is to do some debugging.

When you see an undefined index error, do print_r( $_POST ), or whatever the array is.

The other error is because the mysql connection failed.

[jazmine]

I tried this to check my error:

<?php

$con = mysql_connect("localhost","root","//");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("//", $con);

echo($_POST ['id']);

?>

I get an undefined index error.

Thanks!