我怎么能使AJAX调用没有jQuery?

如何使用JavaScript进行AJAX调用，而不使用jQuery?

当前回答

From youMightNotNeedJquery.com + JSON.stringify

var request = new XMLHttpRequest();
request.open('POST', '/my/url', true);
request.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded; charset=UTF-8');
request.send(JSON.stringify(data));

2017-07-24 15:54:22

其他回答

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。

2014-02-24 20:03:46

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。

2015-08-12 22:01:47

HTML:

<!DOCTYPE html>
    <html>
    <head>
    <script>
    function loadXMLDoc()
    {
    var xmlhttp;
    if (window.XMLHttpRequest)
      {// code for IE7+, Firefox, Chrome, Opera, Safari
      xmlhttp=new XMLHttpRequest();
      }
    else
      {// code for IE6, IE5
      xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
      }
    xmlhttp.onreadystatechange=function()
      {
      if (xmlhttp.readyState==4 && xmlhttp.status==200)
        {
        document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
        }
      }
    xmlhttp.open("GET","1.php?id=99freebies.blogspot.com",true);
    xmlhttp.send();
    }
    </script>
    </head>
    <body>

    <div id="myDiv"><h2>Let AJAX change this text</h2></div>
    <button type="button" onclick="loadXMLDoc()">Change Content</button>

    </body>
    </html>

PHP:

<?php

$id = $_GET[id];
print "$id";

?>

2013-08-06 11:11:46

var load_process = false;
function ajaxCall(param, response) {

 if (load_process == true) {
     return;
 }
 else
 { 
  if (param.async == undefined) {
     param.async = true;
 }
 if (param.async == false) {
         load_process = true;
     }
 var xhr;

 xhr = new XMLHttpRequest();

 if (param.type != "GET") {
     xhr.open(param.type, param.url, true);

     if (param.processData != undefined && param.processData == false && param.contentType != undefined && param.contentType == false) {
     }
     else if (param.contentType != undefined || param.contentType == true) {
         xhr.setRequestHeader('Content-Type', param.contentType);
     }
     else {
         xhr.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
     }


 }
 else {
     xhr.open(param.type, param.url + "?" + obj_param(param.data));
 }

 xhr.onprogress = function (loadTime) {
     if (param.progress != undefined) {
         param.progress({ loaded: loadTime.loaded }, "success");
     }
 }
 xhr.ontimeout = function () {
     this.abort();
     param.success("timeout", "timeout");
     load_process = false;
 };

 xhr.onerror = function () {
     param.error(xhr.responseText, "error");
     load_process = false;
 };

 xhr.onload = function () {
    if (xhr.status === 200) {
         if (param.dataType != undefined && param.dataType == "json") {

             param.success(JSON.parse(xhr.responseText), "success");
         }
         else {
             param.success(JSON.stringify(xhr.responseText), "success");
         }
     }
     else if (xhr.status !== 200) {
         param.error(xhr.responseText, "error");

     }
     load_process = false;
 };
 if (param.data != null || param.data != undefined) {
     if (param.processData != undefined && param.processData == false && param.contentType != undefined && param.contentType == false) {
             xhr.send(param.data);

     }
     else {
             xhr.send(obj_param(param.data));

     }
 }
 else {
         xhr.send();

 }
 if (param.timeout != undefined) {
     xhr.timeout = param.timeout;
 }
 else
{
 xhr.timeout = 20000;
}
 this.abort = function (response) {

     if (XMLHttpRequest != null) {
         xhr.abort();
         load_process = false;
         if (response != undefined) {
             response({ status: "success" });
         }
     }

 }
 }
 }

function obj_param(obj) {
var parts = [];
for (var key in obj) {
    if (obj.hasOwnProperty(key)) {
        parts.push(encodeURIComponent(key) + '=' + encodeURIComponent(obj[key]));
    }
}
return parts.join('&');
}

我的ajax调用

  var my_ajax_call=ajaxCall({
    url: url,
    type: method,
    data: {data:value},
    dataType: 'json',
    async:false,//synchronous request. Default value is true 
    timeout:10000,//default timeout 20000
    progress:function(loadTime,status)
    {
    console.log(loadTime);
     },
    success: function (result, status) {
      console.log(result);
    },
      error :function(result,status)
    {
    console.log(result);
     }
      });

对于中止先前的请求

      my_ajax_call.abort(function(result){
       console.log(result);
       });

2016-12-08 20:20:18

使用下面的代码片段，你可以很容易地完成类似的事情，就像这样:

ajax.get('/test.php', {foo: 'bar'}, function() {});

以下是片段:

var ajax = {};
ajax.x = function () {
    if (typeof XMLHttpRequest !== 'undefined') {
        return new XMLHttpRequest();
    }
    var versions = [
        "MSXML2.XmlHttp.6.0",
        "MSXML2.XmlHttp.5.0",
        "MSXML2.XmlHttp.4.0",
        "MSXML2.XmlHttp.3.0",
        "MSXML2.XmlHttp.2.0",
        "Microsoft.XmlHttp"
    ];

    var xhr;
    for (var i = 0; i < versions.length; i++) {
        try {
            xhr = new ActiveXObject(versions[i]);
            break;
        } catch (e) {
        }
    }
    return xhr;
};

ajax.send = function (url, callback, method, data, async) {
    if (async === undefined) {
        async = true;
    }
    var x = ajax.x();
    x.open(method, url, async);
    x.onreadystatechange = function () {
        if (x.readyState == 4) {
            callback(x.responseText)
        }
    };
    if (method == 'POST') {
        x.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
    }
    x.send(data)
};

ajax.get = function (url, data, callback, async) {
    var query = [];
    for (var key in data) {
        query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
    }
    ajax.send(url + (query.length ? '?' + query.join('&') : ''), callback, 'GET', null, async)
};

ajax.post = function (url, data, callback, async) {
    var query = [];
    for (var key in data) {
        query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
    }
    ajax.send(url, callback, 'POST', query.join('&'), async)
};

2013-08-06 11:16:00

我怎么能使AJAX调用没有jQuery?

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