将参数传递给XMLHttpRequest对象

Question

如何将参数传递给XMLHttpRequest对象？

function setGUID(aGUID) {

    var xhReq = new XMLHttpRequest();

    xhReq.open("POST", "ClientService.svc/REST/SetAGUID" , false);
    xhReq.send(null);
    var serverResponse = JSON.parse(xhReq.responseText);
    alert(serverResponse);
    return serverResponse;
}

我需要使用javascript而不是jquery,在jquery中我得到它使用这个代码,但似乎无法用直接的javascript方式弄清楚它..

function setGUID(aGUID) {

    var applicationData = null;

    $.ajax({
        type: "POST",
        url: "ClientService.svc/REST/SetAGUID",
        contentType: "application/json; charset=utf-8",
        data: JSON.stringify({ aGUID: aGUID }),
        dataType: "json",
        async: false,
        success: function (msg) {

            applicationData = msg;

        },
        error: function (xhr, status, error) { ); }
    });

    return applicationData;

}

Answer 1

互联网上有很多关于"xmlhttprequest post"的教程.我只是复制其中一个:

看一看:

http://www.openjs.com/articles/ajax_xmlhttp_using_post.php

https://www.google.com/search?q=xmlhttprequest+post

var http = new XMLHttpRequest();
var url = "url";
var params = JSON.stringify({ appoverGUID: approverGUID });
http.open("POST", url, true);

http.setRequestHeader("Content-type", "application/json; charset=utf-8");
http.setRequestHeader("Content-length", params.length);
http.setRequestHeader("Connection", "close");

http.onreadystatechange = function() {
    if(http.readyState == 4 && http.status == 200) {
        alert(http.responseText);
    }
}
http.send(params);