Get json data from php and use it in jquery

Question

It is my first time using JSON. I am trying to get JSON data from a php script using ajax.But i am getting this error "Error.Parsing JSON Request failed" "undefined".Help me this is my php script test.php

    $data='{
  "one": "One",
 "two": "Two",
 "three": "Three"
   }';

 header('Content-type: application/json');
 echo json_encode($data);

and here i am getting the data getdata.php

    var sURL = 'http://www.example.com/test.php';   

$.ajax({
            type: "GET",
                            url:sURL,
                            dataType:"jsonp",
                            crossDomain:true,
                            data: {transid:trans_id , bookingdate: dateVal, bookingtime: timeVal, People: peopleVal,affiliateid: affiliate },
                            async: false,
                            contentType:"application/json; charset=utf-8",
                            success: function (data){
                                                                                    var result = JSON.parse(data);
                                            alert(result);
                                            },
                                            error: function (x, e) {

        if (x.status == 0) {
            alert(x.response);
            alert(x + " " + e);
            alert('You are offline!!\n Please Check Your Network.');
        }
        else if (x.status == 404) {
            alert('Requested URL not found.');
        }
        else if (x.status == 500) {
            alert('Internel Server Error.');
        }
        else if (e == 'parsererror') {

            alert('Error.\nParsing JSON Request failed.' + e.statusText);
            alert(x.response);
        } else if (e == 'timeout') {
            alert('Request Time out.');
        } else {
            alert('Unknow Error.\n' + x.responseText);
        }
    }

                            });

This is my first question so excuse any mistakes

GBD · Accepted Answer · 2012-12-15 08:52:00Z

2

As you are using jsonp, add callback function as below

var sURL = 'http://www.example.com/test.php';   

$.ajax({
            type: "GET",
                            url:sURL,
                            dataType:"jsonp",
                            jsonp : "callback",
                            jsonpCallback: "jsonpcallback",
                            crossDomain:true,
                            data: {transid:trans_id , bookingdate: dateVal, bookingtime: timeVal, People: peopleVal,affiliateid: affiliate },
                            async: false,
                            contentType:"application/json; charset=utf-8",
                            success: function (data){
                                                                                    var result = JSON.parse(data);
                                            alert(result);
                                            },
                                            error: function (x, e) {

        if (x.status == 0) {
            alert(x.response);
            alert(x + " " + e);
            alert('You are offline!!\n Please Check Your Network.');
        }
        else if (x.status == 404) {
            alert('Requested URL not found.');
        }
        else if (x.status == 500) {
            alert('Internel Server Error.');
        }
        else if (e == 'parsererror') {

            alert('Error.\nParsing JSON Request failed.' + e.statusText);
            alert(x.response);
        } else if (e == 'timeout') {
            alert('Request Time out.');
        } else {
            alert('Unknow Error.\n' + x.responseText);
        }
    }

 });

function jsonpcallback(rtndata) {
  alert(rtndata.one);
}

In PHP make $data as array and then use json_encode() with return callback.

 $data=array(
  "one"=>"One",
  "two"=>"Two",
  "three"=>"Three"
 );
 header('Content-type: application/json');
 echo $_GET['callback']. '('. json_encode($data) . ')';

edited Dec 15, 2012 at 8:52

answered Dec 15, 2012 at 8:26

GBD

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5 Comments

Karan Tuteja Over a year ago

didnt help.now fire bug is also giving a error of SyntaxError: invalid label [Break On This Error] {"one":"One","two":"Two","three":"Three"}(line 1, col 1)

Bhushan Firake Over a year ago

@KaranTuteja: try myanswer combined with this answer, it should work as json_encode method accepts array only..!!

Karan Tuteja Over a year ago

now its giving this error in firebug SyntaxError: JSON.parse: unexpected character [Break On This Error] var result = JSON.parse(data);

GBD Over a year ago

are you sure want to get json from cross-domain ?

Karan Tuteja Over a year ago

@gbd yes i need cross domain it is very important for my application.

Bhushan Firake · Accepted Answer · 2012-12-15 08:22:27Z

0

Change your

     dataType:"jsonp",

to

dataType:"json",

answered Dec 15, 2012 at 8:22

Bhushan Firake

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2 Comments

Karan Tuteja Over a year ago

tried it didnt help. it doesn't effect anything anyways jsonp is used for cross domain.

Jai Over a year ago

what do you mean by crossdomain(on same proxy)? or from external source?

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