Question

1

JavaScript XMLHttpRequest using JsonP

rated 0 times [ 8] [ 7] / answers: 1 / hits: 41675 / 10 Years ago, tue, april 1, 2014, 12:00:00

I want to send request parameters to other domain

I already know that Cross Scripting needs JsonP and I have used JsonP with Jquery ajax

but i do not figure out how to do Cross Scripting as using XMLHttpRequest

following code my basic XMLHttpRequest code.

i guess i need to chage xhr.setRequestHeader() and i have to add parsing code

please give me any idea

var xhr;

function createXMLHttpRequest(){    

    if(window.AtiveXObject){

        xhr = new ActiveXObject(Microsoft.XMLHTTP);

    }else{

        xhr = new XMLHttpRequest();

    }   

    var url = http://www.helloword.com;   

}



function openRequest(){ 

    createXMLHttpRequest();

    xhr.onreadystatechange = getdata;

    xhr.open(POST,url,true);

    xhr.setRequestHeader(Content-Type,'application/x-www-form-urlencoded');

    xhr.send(data); 

}



function getdata(){

    if(xhr.readyState==4){

        if(xhr.status==200){

            var txt = xhr.responseText;

            alert(txt);

        }

    }   

}

Answers

Only authorized users can answer the question. Please sign in first, or register a free account.

darrylm

Add To Favorites

Follow

Total Points: 499

Total Questions: 131

Total Answers: 108

Location: Saudi Arabia

Member since Mon, Sep 5, 2022

2 Years ago

answered 10 Years ago jocelynkarsynr · Accepted Answer

JSONP does not use XMLHttpRequests.

The reason JSONP is used is to overcome cross-origin restrictions of XHRs.

Instead, the data is retrieved via a script.

function jsonp(url, callback) {

    var callbackName = 'jsonp_callback_' + Math.round(100000 * Math.random());

    window[callbackName] = function(data) {

        delete window[callbackName];

        document.body.removeChild(script);

        callback(data);

    };



    var script = document.createElement('script');

    script.src = url + (url.indexOf('?') >= 0 ? '&' : '?') + 'callback=' + callbackName;

    document.body.appendChild(script);

}



jsonp('http://www.helloword.com', function(data) {

   alert(data);

});

In interest of simplicity, this does not include error handling if the request fails. Use script.onerror if you need that.