johnvj Posted March 5, 2014 Share Posted March 5, 2014 (edited) <html> <body> <form method="post" action=""> Username:<input type="text" name="user"> password:<input type="text" name="pass"> <input type="submit" name="submit" value="submit"> </form> </body> </html> <?php $user=$_POST['user']; $pass=$_POST['pass']; $conn = mysql_connect("localhost","root","root"); if ($conn): $selectDB=mysql_select_db("project",$conn); if (!$selectDB): echo mysql_error(); endif; endif; // how many rows to show per page $rowsPerPage = 2; // by default we show first page $pageNum = 1; // if $_GET['page'] defined, use it as page number if(isset($_GET['page'])) { $pageNum = $_GET['page']; } // counting the offset $offset = ($pageNum - 1) * $rowsPerPage; $query = "SELECT * FROM form1 LIMIT $offset, $rowsPerPage"; $result = mysql_query($query) or die(mysql_error()); if(isset($_POST['submit'])){ if($user=="jishil" && $pass=="1234"){ // print the random numbers echo "FORM 1"; echo "<table width='50%' border='1'>"; echo "<tr><td>USERNAME</td><td>ADDRESS</td><td>COMPANY</td><td>SALARY</td><td>D.O.B</td><td>D.O.J</td></tr>"; while($rows=mysql_fetch_array($result)){ echo "<form action='' method='post'>"; echo "<tr>"; echo "<td>" . "<input type='text' name='Name' value=" . $rows['name']. " </td>"; echo "<td>" . "<input type='text' name='Address' value=" . $rows['address']. " </td>"; echo "<td>" . "<input type='text' name='Comp' value=" . $rows['comp']. " </td>"; echo "<td>" . "<input type='text' name='Sal' value=" . $rows['sal']. " </td>"; echo "<td>" . "<input type='text' name='Birth' value=" . $rows['birth']. " </td>"; echo "<td>" . "<input type='text' name='Join' value=" . $rows['join']. " </td>"; echo "<td>" . "<input type='hidden' name='hidden' value=" . $rows['name']. " </td>"; echo "<td>" . "<input type='submit' name='update' value=update" ." </td>"; echo "</tr>"; echo "</form>"; } echo "</table>"; // how many rows we have in database $query = "SELECT COUNT(name) AS numrows FROM form1"; $result = mysql_query($query) or die('Error, query failed'); $row = mysql_fetch_array($result, MYSQL_ASSOC); $numrows = $row['numrows']; // how many pages we have when using paging? $maxPage = ceil($numrows/$rowsPerPage); // print the link to access each page $self = $_SERVER['PHP_SELF']; $nav = ''; for($page = 1; $page <= $maxPage; $page++) { if ($page == $pageNum) { $nav .= " $page "; // no need to create a link to current page } else { $nav .= " <a href=\"$self?page=$page\">$page</a> "; } } // creating previous and next link // plus the link to go straight to // the first and last page if ($pageNum > 1) { $page = $pageNum - 1; $prev = " <a href=\"$self?page=$page\">[Prev]</a> "; $first = " <a href=\"$self?page=1\">[First Page]</a> "; } else { $prev = ' '; // we're on page one, don't print previous link $first = ' '; // nor the first page link } if ($pageNum < $maxPage) { $page = $pageNum + 1; $next = " <a href=\"$self?page=$page\">[Next]</a> "; $last = " <a href=\"$self?page=$maxPage\">[Last Page]</a> "; } else { $next = ' '; // we're on the last page, don't print next link $last = ' '; // nor the last page link } // print the navigation link echo $first . $prev . $nav . $next . $last; }} ?> Edited March 5, 2014 by requinix please use [code] tags when posting code Quote Link to comment https://forums.phpfreaks.com/topic/286708-why-i-have-to-log-in-again-to-see-pagination-results/ Share on other sites More sharing options...
Ch0cu3r Posted March 5, 2014 Share Posted March 5, 2014 (edited) $_POST data is not remembered during multiple page requests. The post data will only exist when the form is submitted. When you click one of the pagination links, the post data will not exits and so the login form appears. The next page will only display when you filll the login form again. What you should do set a login token in the $_SESSION when you authenticate the user. Then when you go to display the paginated results, check to make sure the login token exists to verify the user is authenticated. if no login token then display the login form. This is how your code should be structured <?php // create a session, so the $_SESSION vars are remembered session_start(); // do the login, when login form is submitted if(isset($_POST['submit'])) { $user=$_POST['user']; $pass=$_POST['pass']; if($user=="jishil" && $pass=="1234") { // set sessions vars when user is authenticated $_SESSION['loggedIn'] = true; // set the loggedIn token to true $_SESSION['username'] = $user; // the users username } } // only display the login form if the user is not authenticated (loggedIn token is not true) if(!isset($_SESSION['loggedIn']) || $_SESSION['loggedIn'] !== true) { // display login form } else { // // user is authenticated display the paginated result } ?> Edited March 5, 2014 by Ch0cu3r Quote Link to comment https://forums.phpfreaks.com/topic/286708-why-i-have-to-log-in-again-to-see-pagination-results/#findComment-1471503 Share on other sites More sharing options...
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