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How to show the data from mysql table for my case?


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i have3 Table

 

block table

id,block

(total 9 blocks based on cityname ex:london,mumbai....)

 

users table

id,username,password,block,access

(this is registration form fields.Select your block from drop down menu)

 

total 3 acess levels

0 for admin,1 for editor and 2 for regular user

 

form table

id,block,scheme,status. (Select your block from drop down menu)

 

i have a working login registration system also successfully get the results from the Form table.

But i need if a user is from london block then show only london block results of Form table and if from mumbai show only mumbai results of Form table.

 

Please help friends

 

my code

<?php
include '../include/functions.php';

if(!loggedin()) {
   header("Location: login.php");
   exit ();
}

$per_page = 3;
$pages_query = mysql_query("SELECT COUNT('id') FROM form");
$pages = ceil(mysql_result($pages_query, 0) / $per_page);

$page = (isset ($_GET['page'])) ? (int)$_GET['page'] : 1;
$start = ($page - 1) * $per_page;

$result = mysql_query("SELECT * FROM form");

echo  "<table>
<tr>
<th>ID</th>
<th>Block</th>
<th>SCHEME NAME</th>
<th>Status</th>
</tr>";
while($row = mysql_fetch_array($result))
 {
   echo "<tr>";
   echo "<td>" . $row['id'] . "</td>";
   echo "<td>" . $row['block'] . "</td>";
   echo "<td>" . $row['scheme'] . "</td>";
   echo "<td>" . $row['status'] . "</td>";


   echo "</tr>";
 }
 echo "</table>";
 echo "Page: ";
 if($pages>=1 && $page<=$pages) {
     for($x=1; $x<=$pages; $x++) {

         echo ($x == $page) ? '<strong><a href="?page='.$x.'">'.$x.'</a></strong>  ' : '<a href="?page='.$x.'">'.$x.'</a>   ' ;
     }
 }

?> 

 

i tried below code but it returns only one row

<?php
include '../include/functions.php';

if(!loggedin()) {
   header("Location: login.php");
   exit ();
}
$user = $_SESSION['username'];
if($user) {
$queryget = mysql_query("SELECT block FROM users WHERE username='$user'") or die("Query is not working");
$row = mysql_fetch_assoc($queryget);

$block = $row['block'];

if($row)
{

$result = mysql_query("SELECT * FROM form WHERE block='$block'");
$row1 = mysql_fetch_array($result);
echo  "<table>
<tr>
<th>ID</th>
<th>Block</th>
<th>SCHEME NAME</th>
<th>Status</th>
</tr>";

   echo "<tr>";
   echo "<td>" . $row1['id'] . "</td>";
   echo "<td>" . $row1['block'] . "</td>";
   echo "<td>" . $row1['scheme'] . "</td>";
   echo "<td>" . $row1['status'] . "</td>";
echo "</tr>";
echo "</table>";
}
}
?> 

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Sounds like you basically need to use your second block of code, but get the block the user wants to view (probably from the url: yourpagename.php?block=test) and then display it:

 

$block = htmlspecialchars($_GET['block'], ENT_QUOTES); //htmlspecialchars to help prevent SQL injection

$queryget = mysql_query("SELECT * FROM form WHERE block='$block'") or die("Query is not working"); // get all columns from the form table where the block is equal to the block set in the URL

 

That should hopefully get you started.

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