referencing drop box selection

Murr

This is probabally the most stupid question in the history of stupid questions and I apologise in advance, but here goes...

Im using embedded PHP in a HTML page. The user will select an option from a drop box, using this selection I will use PHP and MySQL to populate a second drop box. The question I have is how do I find the value of the option they chose in the first place...

I mean, if <select name = "category">

once its submitted how do I reference what was chosen... is it...
$category or
$category->value or
if($selected)

Ive tried searching other forums, tutorials and others but I cant find a clear answer....
Thanks

statrat

the proper way would be $POST['category'] or $GET['category'], though if you have register globals on you can just do $category. you do know you'll have to post the form first to update the second drop down? javascript can do it dynamically on the browser side. just checking.

Murr

OK, bare with me here... this is the code thats on the first page:

<form method = post action = "http://localhost/chooseMovie.php">

<?php
$db = mysql_connect("localhost", "root", "password");

mysql_select_db("archives",$db);

?>
Choose a category:<br>

<select name=category>

<?
$result = mysql_query("SELECT catId, category FROM category",$db);
while ($row = mysql_fetch_object($result))
      {
	echo "<option value=" . $row->catId . ">" . $row->category .  "</option><br>";
      }

mysql_free_result($result);
?>

</select>

This populates the first drop box with the results of a MySQL database.
Then in "chooseMovie.php" I have this code:

<form method = get action = "http://localhost/playMovie.html">

Movies in this category:<br>

<?php
$db = mysql_connect("localhost", "root", "password");

mysql_select_db("archives",$db);

?>
Choose a movie:<br>

<select name=movie>

<?
$result = mysql_query("SELECT movieName FROM category WHERE catId = "$category"",$db);
while ($row = mysql_fetch_object($result))
      {
	echo "<option>" . $row->movieName .  "</option><br>";
      }

mysql_free_result($result);
?>

</select>

the error Im having is with the MySQL query where Im trying to reference what was chosen in the first place....
Hope this is understandable!!
🙁

stolzyboy

<?
$query = "SELECT movieName FROM category WHERE catId = '$category';";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result))
      {
    echo "<option>" . $row['movieName'] .  "</option><br>";
      }

mysql_free_result($result);
?>

Murr

Problem solved, thanks a million!!😃

stolzyboy

just for curiosities' sake, was mine the fix or did you fix it yourself, if you did it what was the fix, and if it was my fix, your welcome

Murr

A bit of both.... the first thing was I had copied and pasted from other code... and had forgotten to change the table name (stoopid mistake!!)....
Secondly, I think the double quots around $category could have been a problem, aswell as forgetting the ';' at the end of the query..... that was your fix..... the final code looks like:

$result = mysql_query("SELECT movieName FROM movies WHERE catId = '$category';",$db);
while ($row = mysql_fetch_object($result))
      {
	echo "<option>" . $row->movieName .  "</option><br>";
      }

mysql_free_result($result);
?>

Thanks again for your help!