PHP array returns null

Question

I have a html form with a button on which there is a onclick event

    <form id = "abc" action = "xyz.php" method = "post">
      <input type="button" name="submitButton" value="Submit" 
            id="submitButton" onclick = "Function()">
     </form>

In the JS code, I am trying to post a value of a textbox and trying to get the array from the PHP through json_encode.

var valueToSend = $('#textbox').val();
var imagesReturned = new Array();
$.ajax({
        url: "xyz.php",
        type: "POST",
        data: valueToSend,
        dataType:"json",
        success: function(result) {        
            var data = jQuery.parseJSON(result);
            alert(data);
        }
    });

And here is my PHP code,

        if(isset($_POST['valueToSend'])) 
        {
            $searchValues = explode(' ',$_POST['valueToSend']);
           // Some processing here
            echo (json_encode($responseArray));

        }

When I alert the data, I get null, What am I missing?

Thanks!

Its an array of strings.. For example, it contains abc,xyz,pqr etc. — user1821499
– user1821499, Commented Apr 16, 2013 at 17:49
Add a breakpoint on the line that you use parseJSON and see if result is the expected value. — Ricardo Lohmann
– Ricardo Lohmann, Commented Apr 16, 2013 at 17:50
You don't need var data = jQuery.parseJSON(result); in your callback. It's already parsed. On the other hand, your JSON response from PHP will always be empty. You'll never send the proper response because it'll contain invalid data. — N.B.
– N.B., Commented Apr 16, 2013 at 17:51

Joshua Burns · Accepted Answer · 2013-04-16 18:24:31Z

3

The problem is within your Ajax request, the data key expects a key/value pair, not just a value.

What you're looking for is something similar to the following:

var valueToSend = {
  'valueToSend': $('#textbox').val()
}

In addition, you're setting the dataType key to json, so there is no need to pass the data variable through parseJSON. As such, the success function only needs to be declared as follows:

'success': function(result) {
  alert(result);
}

edited Apr 16, 2013 at 18:24

answered Apr 16, 2013 at 17:50

Joshua Burns

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2 Comments

Mike Brant Over a year ago

Also note that since you are setting dataType:"json", you don't need to parse the response, jQuery will do that for you such that you can work with the result object directly.

Joshua Burns Over a year ago

you're absolutely right, i'll update. thanks for catching that :)

웃웃웃웃웃 · Accepted Answer · 2013-04-16 17:50:31Z

1

var valueToSend = $('#textbox').val();
var imagesReturned = new Array();
$.ajax({
        url: "xyz.php",
        type: "POST",
        data: {"valueToSend":valueToSend},
        dataType:"json",
        success: function(result) {        
            var data = jQuery.parseJSON(result);
            alert(data);
        }
    });

correct the format of data in ajax

answered Apr 16, 2013 at 17:50

웃웃웃웃웃

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1 Comment

user1821499 Over a year ago

If I change data to data: {valueToSend:valueToSend}, The functions returns error, it doesn't go in the success function.

Manoj Purohit · Accepted Answer · 2013-04-16 19:18:14Z

1

data field of yours look suspicious to me

   $.ajax({
            url: "xyz.php",
            type: "POST",
            data:{valueToSend:valueToSend},
            dataType:"json",
            success: function(result) {        
                alert(result);
            }
        });

edited Apr 16, 2013 at 19:18

answered Apr 16, 2013 at 17:50

Manoj Purohit

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2 Comments

user1821499 Over a year ago

Unfortunately, changing the data field doesn't work, it goes inside the error than success function, if i send key : value pair.. If I send just the value its success with null response!

user1821499 Over a year ago

It does not alert anything, If I add code for Error : function, It goes inside that.

user1821499 · Accepted Answer · 2013-04-16 20:25:02Z

0

The problem was in the PHP code.

I removed the check, if(isset($_POST['valueToSend'])) and its working...

answered Apr 16, 2013 at 20:25

user1821499

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