jQuery json from PHP result

Question

I have this jQuery script

var dataString = "class_id="+class_id;

$.ajax({
    type: "POST",
    url: "page.php",
    data: dataString,
    success: function (msg) {
        //stuck here
    },
    error: function () {
        showNotification("error", "Could not process at this time, try again later."); //this is a function created by me (which works fine so I just left the code in here)
    }
});

my PHP output is something like this

echo '{status:1,message:"Success"}';

or

echo '{status:0,message:"Failure"}';

what I am trying to do in jQuery success: function(...) part is check if status is 0 or 1 and then show the message.

I tried to do is

success: function(text) {
   if(parseInt(text.status) == 1) {
      alert(text.message); // this is the success, the status is 1
   } else {
      alert(text.message); // this is the failure since the status is not 1
   }
}

which didn't work, it was only outputing the else statement, even though the status was 1

Sorry, forgot to post about it, I'll edit it with that info. — Grigor
– Grigor, Commented Apr 6, 2012 at 17:55

Quentin · Accepted Answer · 2012-04-06 17:56:13Z

3

Your PHP is generating invalid JSON, and shows no sign of setting an appropriate content type header to tell the browser to treat it as JSON in the first place. So first, fix the PHP:

header('application/json');
echo json_encode(Array("status" => 1, "message" => "Success"));

Then:

success: function (msg) {
    alert(msg.message)
},

answered Apr 6, 2012 at 17:56

Quentin

949k136 gold badges1.3k silver badges1.4k bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Grigor Over a year ago

the problem was unspecified dataType

Quentin Over a year ago

@Grigor — dataType is how you tell jQuery to ignore the server's content-type header and treat the data as something else. Setting the correct content-type is the better option.

Grigor Over a year ago

I tried setting the content-type but it wouldn't work still, and when I set the dataType it worked fine.

The Alpha · Accepted Answer · 2012-04-06 18:00:35Z

2

You can also use

PHP

echo json_encode(Array("status" => 1, "message" => "Success"));

JS

Inside your call back function use

success: function (msg) {
    $.parseJSON(msg);
    alert(msg.message);
}

The parseJSON will convert the json string returned/echoed by PHP in to json object.

answered Apr 6, 2012 at 18:00

The Alpha

147k30 gold badges294 silver badges313 bronze badges

Comments

Selvakumar Arumugam · Accepted Answer · 2012-04-06 17:55:19Z

1

Try something like below,

$.ajax({
    type: "POST",
    url: "page.php",
    data: dataString,
    dataType: 'json',
    success: function (msg) {
        if (msg.status == 0) {
          alert("Success " + msg.message);
        } else if (msg.status == 1) {
          alert("Error " + msg.message);
        }
    },
    error: function () {
        showNotification("error", "Could not process at this time, try again later."); //this is a function created by me (which works fine so I just left the code in here)
    }
});

answered Apr 6, 2012 at 17:55

Selvakumar Arumugam

79.9k15 gold badges123 silver badges139 bronze badges

Comments

kappa · Accepted Answer · 2012-04-06 18:01:40Z

1

If you don't specify in $.ajax the type 'json' data passed to response handler is treated as string. While if you specify 'json' dataType parameter you can use:

msg.status

and

msg.message

As a hint i suggest in php to use the json_encode function to generate json output.

answered Apr 6, 2012 at 18:01

kappa

1,5798 silver badges19 bronze badges

Collectives™ on Stack Overflow

jQuery json from PHP result

4 Answers 4

3 Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

3 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related