[RESOLVED] Got the dropdown menu populated from MySQL, but not the results after submit

suttercain

I have a dropdown menu populated from the MySQL database in a form. When I hit submit I want the results of the selection to be displayed... but I can't get it.

FORM CODE:

<form action="comic_index_view.php" method="post">
				  <?php 
				  $result = mysql_query("SELECT title FROM comics GROUP BY title") or die(mysql_error());
 				  $options="";
				  echo "<select>";
 				  while($row = mysql_fetch_array($result)){
 				  $title=$row["title"];
 				  echo "<option value'" . $title . "'>".$title.'</option>';
 					}
				  echo "</select>";
				  ?>
				  <input type="submit" name="submit">
				  </form>

RESULTS PAGE:

<?php
require('get_connected.php');

$title = $_POST['title'];
$sql = mysql_query("SELECT title, issue_number, cover_date, comic_id FROM comics WHERE title ='" . $title . "'") or die(mysql_error());
	if ($result = mysql_query($sql)) {
      if (mysql_num_rows($result)) {
        while ($row = mysql_fetch_assoc($result)) {
		echo $row['title'];
		}
      }
    }

echo $row['title'] . "<br>";
?>

I know this is a common thing people do with PHP but I could not find any info on the web. Can any one suggest a possible solution?

Thank you.

NogDog

You need to "name" your select element:

echo "<select name='title'>";

suttercain

I had tired that before and it didn't work but I tried it again...

<?php 
				  $result = mysql_query("SELECT title FROM comics GROUP BY title") or die(mysql_error());
 				  $options="";
				  echo "<select name='title'>";
 				  while($row = mysql_fetch_array($result)){
 				  $title=$row["title"];
 				  echo "<option value='" . $title . "'>".$title.'</option>';
 					}
				  echo "</select>";
				  ?>
				  <input type="submit" name="submit">
				  </form>

I also noticed in my first code I didn't have '=' before value. I ran the code and it still didn't work. Any other suggestions? Thanks.

bastien

if you want a re-usable function, i have this here