Javascript 将参数传递给 XMLHttpRequest 对象

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时间:2020-08-23 10:59:55  来源:igfitidea点击:

Pass Parameters To XMLHttpRequest Object

javascript

提问by Nick LaMarca

How can I pass parameters to the XMLHttpRequest Object?

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;
}

I need to use javascript instead of jquery, in jquery I got it to work with this code, but cant seem to figure it out the straight javascript way..

我需要使用 javascript 而不是 jquery,在 jquery 中我让它与这段代码一起工作,但似乎无法弄清楚直接的 javascript 方式..

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;

}

回答by lolol

There's a lot of tutorials about "xmlhttprequest post" on the internet. I just copy one of then:

网上有很多关于“xmlhttprequest post”的教程。我只是复制其中之一:

Take a look:

看一看:

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

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

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

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);

回答by mohamad khaled

Just remove these two lines (which are not allowed to set these headers):

只需删除这两行(不允许设置这些标题):

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