Returning arrays from php with ajax and json

Question

-I have an array that contains more arrays in my php file. I am sending a integer variable (Possibly something else for sometime later that isn't the problem)

-I want php file to return array according to the incoming variable (in this example just 1 and 2.)

-I tried somethings from other posts but they were too complicated and I couldn't understand any of them. And I want to be able to change it whenever I want.

This is what I have done so far..

index.php

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script type="text/javascript" src="jquery-2.1.4.min.js"></script>
<script type="text/javascript">     
var sayac = 1;
$(document).ready(function() {

    $(document).delegate("#klikme", "click", function() {
        $.post('sayac.php',{"komut": sayac }, function(data){

            var res = $.parseJSON(data);
            $('#output').html(res.result[0]);
            $('#output2').html(res.result[1]);
        });
        sayac = sayac + 1;
        if(sayac > 2)
            sayac = 1;
    });
});
</script>
<div id = "klikme">
    KLİK ME!
</div>

<div id = "output">
<?php include 'sayac.php'?>

</div>

<div id = "output2">

</div>

sayac.php

<?php 

    $returnThis = array( array("One" => "Sample Stuff Bla Bla this is one!","Two" => "Sample Text One"), 
                        array("One" => "Sample Stuff Bla Bla this is one!", "Two"=> "Sample Text Two"));

    if(isset($_POST["komut"]))
    {
        switch($_POST["komut"])
        {
            case 1:
                echo json_encode($returnThis[0]) ;
                break;
            case 2:
                echo json_encode($returnThis[1]); 
                break;
            default:
                echo "Something is wrong";
                break;
        }
    }

?>

I want to return array and in my index I want to fill each of my output div areas with returned array's indexes but these doesn't work

That was emberessing... Fixed it. Didn't work. I still can't get any response. — Çağatay Sarp Aksu
– Çağatay Sarp Aksu, Commented Jun 28, 2015 at 15:01
The problem is in the PHP or the JS? You should focus your question on the problematic issue — Michael Ben-Nes
– Michael Ben-Nes, Commented Jun 28, 2015 at 15:08

Death-is-the-real-truth · Accepted Answer · 2015-06-28 15:09:55Z

1

Just change your stuff like below:-

<script type="text/javascript">     

$(document).ready(function() {
var sayac = 1;
    $(document).delegate("#klikme", "click", function() {
        $.post('sayac.php',{"komut": sayac }, function(data){

            var res = $.parseJSON(data);
            console.log(res); // since response is an object so you cannot access in the way that you did.
            $('#output').html(res.One); // get first value
            $('#output2').html(res.Two); // get second value
        });
        sayac = sayac + 1;
        if(sayac > 2)
            sayac = 1;
    });
});
</script>

answered Jun 28, 2015 at 15:09

Death-is-the-real-truth

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Comments

Allkin · Accepted Answer · 2015-06-28 15:25:27Z

1

Edit the javascript code like this:

$(document).ready(function() {

$('#klikme').click(function() {
    $.ajax({
        type: 'POST',
        url: 'sayac.php',
        data: 'komut='+sayac,
        dataType: 'json',
        cache: false,
        success: function(result) {
        $('#output').html(result.One);
        $('#output2').html(result.Two);
        }
    });
});

});

answered Jun 28, 2015 at 15:25

Allkin

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1 Comment

Death-is-the-real-truth Over a year ago

your code is correct but , what is the issue using $.post. this is latest and fast than $.ajax.

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