Question

In my admin section, I want the user to be able to click a button and a HTML report to display all fields from all tables with the user id that is set in that session.
So $user_id = intval($_SESSION['user_id']);

Would the button be something like:

<?php
if (isset($_POST['doReport'])) {

$link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Couldn't make connection.");

$query = ("SELECT user_id FROM users WHERE user_id = '$user_id');
}
// redirect to report page
            header('Location: report.php?id=' . $user_id);
?>
 <p align="center">
                <input name="doReport" type="submit" id="doThesisReport" value="View Report">

Also, what would be the best way to lay it out in the report page as I have 7 tables.
I am looking for a quick easy fix
Thank you!

Answer 1

The good way to achieve this would be :

• Make a button witch redirects to a new page, say “report.php”.

• Now create the following logic for redirect page:

  1. Fetch the USER-ID from session using the code you’ve used above.
  2. Make the database connection on this page.
  3. You forgot to use mysql_select_db(); to select the database.

  4. Write the query like :

     $query = ("SELECT user_id FROM users WHERE user_id = '$user_id'");
     // you left "
     

  5. Fire the query using $result=mysql_query($query); Note that the result from database is saved into $result.
  6. You can use tables to display data as per you liking but quick way would be to use print_r($result)
  7. Close the connection using mysql_close();

Hopefully since the question is not clear, I’ve posted the basic logic.

EDITED : forgot to put query in code block.