Sending back multiple results from PHP via AJAX

Question

i have setup some ajax, which i am just testing now, pretty much the idea behind it is to send some data from dropdown boxes to a php script, do some calculations and then return the result, it does it well and returns the result, but now rather than just sending back one result and outputting that, i want to send back multiple results and output them, i am able to send multiple data to the php script, so i am sure i can send multiple back.

Anyway it only sends the first result back and not the rest.

Here is the AJAX

 <script>
$("document").ready(function (){ 

    $(".add_extension").change(function(){


        var m = document.getElementById('meter_square');
        var meter_square = m.options[m.selectedIndex].value;

        var s = document.getElementById('story_height');
        var story_height = s.options[s.selectedIndex].value;

     $.ajax({
            type: "GET",
            url: "script.php",
            data: { meter_square: meter_square, story_height: story_height },
            dataType: "json",
            statusCode: {
                200: function (result, result2)
                {
                    $("#expected_gain").html(result.value);
                $("#house_price").html(result2.value2);
                }

            }
        });
})
});
</script>

And here is the php script

    <?php 

$meter_square = $_GET["meter_square"];
$story_height = $_GET["story_height"];


$result = $meter_square + $story_height;
$result2 = $meter_square * $story_height;

echo json_encode(array("value" => $result, "value2" => $result2));

 ?>

You can see that i have already tried to give it a go from what i thought might work, if you need any other code or want me to remove the code i added which doesn't work, then let me know.

Thanks for all and any help

Why do you need to do addition and multiplication in PHP? JS is perfectly capable — Martin
– Martin, Commented Jul 21, 2012 at 16:14
i am just using it as a testing part to get the ajax to work first, the php script is going to be pretty big, so i am doing that after — Al Hennessey
– Al Hennessey, Commented Jul 21, 2012 at 16:17

Waleed Khan · Accepted Answer · 2012-07-21 16:15:04Z

7

You're only going to receive one response object:

function (response) {
    $("#expected_gain").html(response.value);
    $("#house_price").html(response.value2);
}

answered Jul 21, 2012 at 16:15

Waleed Khan

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2 Comments

Al Hennessey Over a year ago

brilliant, that works perfectly, will this work if i want to send back 6 different results?

Waleed Khan Over a year ago

It will work more or less the same: if you want to send back, for example, array( "width" => 10, "height" => 5, "depth" => 3 ), you'll be able to use response.width, response.height, and response.depth.

Pavel K · Accepted Answer · 2012-07-21 16:21:17Z

4

Try this. Think it will help. No need to use status codes if u gonna use only success case

 $.ajax({
        type: "GET",
        url: "script.php",
        data: { meter_square: meter_square, story_height: story_height },
        dataType: "json",
        success: function(data){
            $("#expected_gain").html(data.value);
            $("#house_price").html(data.value2);
        }
    });

answered Jul 21, 2012 at 16:21

Pavel K

4963 silver badges17 bronze badges

1 Comment

Linuxios Over a year ago

Good catch with the success instead of the status code. +1!

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