Get php return value using ajax

Question

I am having a form which gets value from the user and stores it to the database. On submitting the form ,it calls the action.php file using ajax call.

e.preventDefault();

$.ajax({

    type: "POST",
    url: "action.php",
    data: senData,
    dataType: "JSON",

    success: function(data) {

        $("#name").val("");
        $('.msg').fadeIn(500);
        $('.msg').text("" + data.result + "");

    }

});

The values are stored in the database without any errors, but I want to display a notification to the user after submitting the form inside the msg div.

In my action.php file I have added a JSON Encode statement to return a message too.

$msg = 'Thanks Yo Yo';
echo json_encode(array("result" => $msg));

But it is not working i.e, when I submit the form, it stores the data to the database and the webpage refreshes itself without displaying any message inside the .msg div.

Am I doing something wrong and is there a better way to do it??

can you post your complete function, which is making a ajax call? — jaysingkar
– jaysingkar, Commented Aug 7, 2016 at 5:10
There is no problem with the HTML code or AJAX function, as the data get perfectly stored in the database. I just don't want the page to be refreshed and to display a return message from the PHP inside the msg div. — It Assistors
– It Assistors, Commented Aug 7, 2016 at 5:20
Have u tried adding alert(data) inside success in ajax call? — coder
– coder, Commented Aug 7, 2016 at 5:29

Duncan · Accepted Answer · 2016-08-07 05:31:08Z

1

You need to parse the JSON when it is returned to your javascript.

// Parse the response to JSON
var res = JSON.Parse(data);
$('.msg').text(res.result);

Your code should look like this.

e.preventDefault();

$.ajax({
 type: "POST",
 url: "action.php",
 data: senData,
 dataType: "JSON",

 success: function(data) {
    var res = JSON.Parse(data);
    $("#name").val("");
    $('.msg').fadeIn(500).text(res.result);

 }

});

answered Aug 7, 2016 at 5:31

Duncan

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