如何在nodejs中创建xhr soap请求？

Question

如何在Node.js中创建xhr请求？这是我的js文件：

var XMLHttpRequest = require("xmlhttprequest").XMLHttpRequest;

var url="https://xx.xxx.xxx.xxx/server";

function soap() {
    var xmlhttp = new XMLHttpRequest();
    xmlhttp.open('POST', url, true);
    xmlhttp.setRequestHeader("SOAPAction", "https://xx.xxx.xxx.xxx/server/myLoginProcess");

    // build SOAP request   
    var sr =
'<?xml version="1.0" encoding="UTF-8"?>'+
'<SOAP-ENV:Envelope xmlns:SOAP-ENV=http://schemas.xmlsoap.org/soap/envelope/'+
    'xmlns:ns1=https://xx.xxx.xxx.xxx/server/'+
    'xmlns:xsd=http://www.w3.org/2001/XMLSchema/'+
    'xmlns:SOAP-ENC=http://schemas.xmlsoap.org/soap/encoding/'+
    'SOAP-ENV:encodingStyle=http://schemas.xmlsoap-org/encoding/>'+
    '<SOAP-ENV:Body>'+
        '<ns1:ihLoginProcess>'+
            '<sEmail xsi:type="xsd:string">me@mail.de</sEmail>'+
            '<sPasswort xsi:type="xsd:string">password</sPasswort>'+
            '<sDeviceID xsi:type="xsd:string">00-14-22-01-23-45</sDeviceID>'+
        '</ns1:ihLoginProcess>'+
    '</SOAP-ENV:Body>'+
'</SOAP-ENV:Envelope>';

        xmlhttp.onreadystatechange = function () 
        {
            console.log("onreadystatechange ");
            if (xmlhttp.readyState == 4) {
                if (xmlhttp.status == 200) {
                    console.log("status 200");
                    console.log('done !');

                }
            }else if (xmlhttp.readyState == 1){
                console.log("readyState 1");
            }
        }
        // Send the POST request
        xmlhttp.setRequestHeader('Access-Control-Allow-Origin', '*');
        xmlhttp.setRequestHeader('Content-Type', 'text/xml;; charset="utf-8"');
        xmlhttp.send(sr);
        // send request
        // ...
}       
soap();

在我写的终端中： 节点myJSFile.js

结果是：

onreadystatechange
readyState 1
onreadystatechange

但它应该是readyState 4并且应该将sr发送到服务器。因此，我应该返回一个XML字符串。