manjumadhav93
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Hi experts,I have the following problem.I have a file login.php which is used to allow users to log in.When the form is submitted it is redirected to checkuserlogin.php.If the credentials are right he is redirected to main.php.But in main.php if the user clicks back browser button,he is redirected to login.php and in login.php if he presses forward button,he is redirected to main.php.Can any one tell he how to make sure the application is not redirected to main.php when user presses forward button and how to make sure the application is not redirected to login.php when user presses back button.
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Sorry for the beginner question, here I'm trying to retrieve data from the database and display it in the table format. But only table headers are printed, and not the actual values. The count variable is echoing 2 saying that data is present and correctly retrieved. Can anyone help? <?php include 'connect.php'; error_reporting(E_ALL ^ E_DEPRECATED); error_reporting(E_ERROR | E_PARSE); $sql="SELECT * FROM `resources` as r INNER JOIN `project_resources` as pr ON r.res_id =pr.res_id WHERE project_id='$_POST[project_id]'"; $result=mysql_query($sql); $count=mysql_num_rows($result); if($result === FALSE) { die(mysql_error()); } echo "$count"; echo '<table> <tr> <th>Resource ID</th> <th>Resource Name</th> <th>Email</th> <th>Phone Number</th> <th>Reporting Manager</th> <th>Role</th> <th>Designation</th> </tr>'; while ($row = mysql_fetch_array($result)) { echo ' <tr> <td>'.$row['res_id'].'</td> <td>'.$row['res_name'].'</td> <td>'.$row['email'].'</td> <td>'.$row['phone_number'].'</td> <td>'.$row['reporting_manager'].'</td> <td>'.$row['role'].'</td> <td>'.$row['designation'].'</td> </tr>'; } echo ' </table>'; ?>
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As a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html>