Goon Posted July 9, 2008 Share Posted July 9, 2008 Hi, I've created a query which retrieves all data from my mysql table and displays the results on a new page. Can anyone tell me how to display the results on the same page? The code used can be seen below. Also, can I edit the code to return data from the database based on the what the user selects (from a dropdown menu) on the form rather then returning all data from the table. Any help would be much appreciated Code: $result = mysql_query("SELECT * FROM $table") or die(mysql_error()); echo "<table border='1'>"; echo "<tr> <th>Vehicle</th> <th>Size</th> </tr>"; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo "<tr><td>"; echo $row['Vehicle']; echo "</td><td>"; echo $row['Price']; echo "</td></tr>"; } echo "</table>"; ?> Quote Link to comment Share on other sites More sharing options...
Wolphie Posted July 9, 2008 Share Posted July 9, 2008 Please use the Code BB tags in future. But the answer to your question is to submit the form to itself using $_SERVER['PHP_SELF'] For example: <?php if(isset($_POST['submit'])) { if(!empty($_POST['name']) { echo $_POST['name']; } else { echo 'Please enter your name!'; } } ?> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST"> <p>Name: <input type="text" name="name" /></p> <p><input type="submit" name="submit" value="Submit" /></p> </form> Quote Link to comment Share on other sites More sharing options...
Goon Posted July 9, 2008 Author Share Posted July 9, 2008 Thanks Wolphie Quote Link to comment Share on other sites More sharing options...
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