Question

First Id like to thank everyone for such positive response on my previous question.

Now I got another question I need help with.

I have a dropdown menu with a list of items. The list is generated inside while loop. Here is the code:

$query = "SELECT Key, Short FROM product WHERE Active = 1 OR Short LIKE 'Blue%'";
$run = mysql_query($query) or die(mysql_error());

echo 'Product: <br />';
?>

<select id="select2" name="select2">
<?php
$ids = 0;
echo "<option selected='selected'>-Select product-</option>";
while($rows = mysql_fetch_assoc($run)) {
    echo "<option value=$ids>".$rows['Short']."</option>";
    $ids++;
}
?>
</select><br /><br />

Now what I need to do is create another dropdown menu below this one and show contracts depending on which option they selected from dropdown menu. Each item they select also have a number called Key. Now inside another table called contracts I have stored all contracts with the same value Key. So…in the second dropdown menu I have to show the contracts based on the key they selected with the item in the first dropdown menu.

I really hope it is clear enough to understand, I am a little confused.

Update: Ok, here is new code:

index.php

$("select#select2").change(function(){
    $.ajax({
        type: "GET",
        url: "process.php",
        data: "selected_key=" + $(this).val(),
        success: function(result) {
            $("select#text2").html(result);
        }
    });
});

</script>

<select id="text2" name="text2">

</select>

And here is my process.php

<?php ## URL_TO_GET_CONTRACTS_FOR_KEY ##
$selectedKey = $_GET['selected_key'];
$query = "SELECT * FROM contacts WHERE Key = '".$selectedKey."'";
$run = mysql_query($query);
while($row = mysql_fetch_assoc($run)) {
    echo "<option value='..'>..</option>";
    } ?>

But I cant see anything displayed in my text2 dropdown menu.

Answer 1

Think you best use option value $rows[‘Key’] in first dropdown menu and add an jQuery selector that takes the selected value and gets te corresponding contracts for that key.

Should be something like this..

PHP CODE

$query = "SELECT Key, Short FROM product WHERE Active = 1 OR Short LIKE 'Blue%'";
$run = mysql_query($query) or die(mysql_error());
echo 'Product: <br />'; ?>
<select id="select2" name="select2">
<?php
echo "<option selected='selected'>-Select product-</option>";
while($rows = mysql_fetch_assoc($run)) {
    echo "<option value='".$rows['Key']."'>".$rows['Short']."</option>";
}
?>
</select>

jQUERY CODE

$("select#select2").change(function(){
    $.ajax({
        type: "GET",
        url: "URL_TO_GET_CONTRACTS_FOR_KEY",
        data: "selected_key=" + $(this).val(),
        success: function(result) {
            $("select#NEWSELECT").html(result);
        }
    });
});

Explanation:
URL_TO_GET_CONTRACTS_FOR_KEY is an url to a PHP file you have to write. In that file you have access to $_GET[‘selected_key’], use that value to get the contracts for that key. In that file you should echo the “” tags for the second select. Like this:

<?php ## URL_TO_GET_CONTRACTS_FOR_KEY ##
$selectedKey = $_GET['selected_key'];
$query = ..
$run = ..
while($row = mysql_fetch_assoc($run)) {
    echo "<option value='..'>..</option>";
} ?>

NEWSELECT should be replaced by the id of the select where the returned options should be placed in.