如何使用JavaScript进行AJAX调用,而不使用jQuery?


当前回答

xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
        alert(this.responseText);
    }
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();

其他回答

这里有一个非常好的纯javascript解决方案

/*create an XMLHttpRequest object*/

let GethttpRequest=function(){  
  let httpRequest=false;
  if(window.XMLHttpRequest){
    httpRequest   =new XMLHttpRequest();
    if(httpRequest.overrideMimeType){
    httpRequest.overrideMimeType('text/xml');
    }
  }else if(window.ActiveXObject){
    try{httpRequest   =new ActiveXObject("Msxml2.XMLHTTP");
  }catch(e){
      try{
        httpRequest   =new ActiveXObject("Microsoft.XMLHTTP");
      }catch(e){}
    }
  }
  if(!httpRequest){return 0;}
  return httpRequest;
}

  /*Defining a function to make the request every time when it is needed*/

  function MakeRequest(){

    let uriPost       ="myURL";
    let xhrPost       =GethttpRequest();
    let fdPost        =new FormData();
    let date          =new Date();

    /*data to be sent on server*/
    let data          = { 
                        "name"      :"name",
                        "lName"     :"lName",
                        "phone"     :"phone",
                        "key"       :"key",
                        "password"  :"date"
                      };

    let JSONdata =JSON.stringify(data);             
    fdPost.append("data",JSONdata);
    xhrPost.open("POST" ,uriPost, true);
    xhrPost.timeout = 9000;/*the time you need to quit the request if it is not completed*/
    xhrPost.onloadstart = function (){
      /*do something*/
    };
    xhrPost.onload      = function (){
      /*do something*/
    };
    xhrPost.onloadend   = function (){
      /*do something*/
    }
    xhrPost.onprogress  =function(){
      /*do something*/
    }

    xhrPost.onreadystatechange =function(){

      if(xhrPost.readyState < 4){

      }else if(xhrPost.readyState === 4){

        if(xhrPost.status === 200){

          /*request succesfull*/

        }else if(xhrPost.status !==200){

          /*request failled*/

        }

      }


   }
  xhrPost.ontimeout = function (e){
    /*you can stop the request*/
  }
  xhrPost.onerror = function (){
    /*you can try again the request*/
  };
  xhrPost.onabort = function (){
    /*you can try again the request*/
  };
  xhrPost.overrideMimeType("text/plain; charset=x-user-defined-binary");
  xhrPost.setRequestHeader("Content-disposition", "form-data");
  xhrPost.setRequestHeader("X-Requested-With","xmlhttprequest");
  xhrPost.send(fdPost);
}

/*PHP side
<?php
  //check if the variable $_POST["data"] exists isset() && !empty()
  $data        =$_POST["data"];
  $decodedData =json_decode($_POST["data"]);
  //show a single item from the form
  echo $decodedData->name;

?>
*/

/*Usage*/
MakeRequest();

这可能会有帮助:

function doAjax(url, callback) {
    var xmlhttp = window.XMLHttpRequest ? new XMLHttpRequest() : new ActiveXObject("Microsoft.XMLHTTP");

    xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
            callback(xmlhttp.responseText);
        }
    }

    xmlhttp.open("GET", url, true);
    xmlhttp.send();
}

下面的几个例子的一个小组合,创造了这个简单的作品:

function ajax(url, method, data, async)
{
    method = typeof method !== 'undefined' ? method : 'GET';
    async = typeof async !== 'undefined' ? async : false;

    if (window.XMLHttpRequest)
    {
        var xhReq = new XMLHttpRequest();
    }
    else
    {
        var xhReq = new ActiveXObject("Microsoft.XMLHTTP");
    }


    if (method == 'POST')
    {
        xhReq.open(method, url, async);
        xhReq.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
        xhReq.setRequestHeader("X-Requested-With", "XMLHttpRequest");
        xhReq.send(data);
    }
    else
    {
        if(typeof data !== 'undefined' && data !== null)
        {
            url = url+'?'+data;
        }
        xhReq.open(method, url, async);
        xhReq.setRequestHeader("X-Requested-With", "XMLHttpRequest");
        xhReq.send(null);
    }
    //var serverResponse = xhReq.responseText;
    //alert(serverResponse);
}

// Example usage below (using a string query):

ajax('http://www.google.com');
ajax('http://www.google.com', 'POST', 'q=test');

或者如果你的参数是object(s) -轻微的额外代码调整:

var parameters = {
    q: 'test'
}

var query = [];
for (var key in parameters)
{
    query.push(encodeURIComponent(key) + '=' + encodeURIComponent(parameters[key]));
}

ajax('http://www.google.com', 'POST', query.join('&'));

两者都应该完全兼容浏览器+版本。

现在在现代浏览器中有一个更好的本机Fetch API可用。fetch()方法允许您发出web请求。 例如,从/get-data请求一些JSON:

let options = {
  method: 'GET',      
  headers: {}
};

fetch('/get-data', options)
.then(response => response.json())
.then(body => {
  // Do something with body
});

更多细节请参见MDN Web Docs: Using Fetch API。

这是一个没有JQuery的JSFiffle

http://jsfiddle.net/rimian/jurwre07/

function loadXMLDoc() {
    var xmlhttp = new XMLHttpRequest();
    var url = 'http://echo.jsontest.com/key/value/one/two';

    xmlhttp.onreadystatechange = function () {
        if (xmlhttp.readyState == XMLHttpRequest.DONE) {
            if (xmlhttp.status == 200) {
                document.getElementById("myDiv").innerHTML = xmlhttp.responseText;
            } else if (xmlhttp.status == 400) {
                console.log('There was an error 400');
            } else {
                console.log('something else other than 200 was returned');
            }
        }
    };

    xmlhttp.open("GET", url, true);
    xmlhttp.send();
};

loadXMLDoc();