cool Posted November 18, 2011 Report Share Posted November 18, 2011 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>"; } } ?> Quote Link to comment Share on other sites More sharing options...
falkencreative Posted November 19, 2011 Report Share Posted November 19, 2011 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. Quote Link to comment Share on other sites More sharing options...
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