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

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

其他回答

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>
  <script>
    var xmlDoc = null ;

  function load() {
    if (typeof window.ActiveXObject != 'undefined' ) {
      xmlDoc = new ActiveXObject("Microsoft.XMLHTTP");
      xmlDoc.onreadystatechange = process ;
    }
    else {
      xmlDoc = new XMLHttpRequest();
      xmlDoc.onload = process ;
    }
    xmlDoc.open( "GET", "background.html", true );
    xmlDoc.send( null );
  }

  function process() {
    if ( xmlDoc.readyState != 4 ) return ;
    document.getElementById("output").value = xmlDoc.responseText ;
  }

  function empty() {
    document.getElementById("output").value = '<empty>' ;
  }
</script>

<body>
  <textarea id="output" cols='70' rows='40'><empty></textarea>
  <br></br>
  <button onclick="load()">Load</button> &nbsp;
  <button onclick="empty()">Clear</button>
</body>
</html>

2011-12-19 20:31:20

老了，但我会尝试，也许有人会发现这个信息有用。

这是执行GET请求并获取一些JSON格式数据所需的最小代码量。这只适用于现代浏览器，如最新版本的Chrome, FF, Safari, Opera和Microsoft Edge。

const xhr = new XMLHttpRequest();
xhr.open('GET', 'https://example.com/data.json'); // by default async 
xhr.responseType = 'json'; // in which format you expect the response to be


xhr.onload = function() {
  if(this.status == 200) {// onload called even on 404 etc so check the status
   console.log(this.response); // No need for JSON.parse()
  }
};

xhr.onerror = function() {
  // error 
};


xhr.send();

还可以查看新的Fetch API，它是XMLHttpRequest API的基于承诺的替代品。

2015-09-27 06:20:39

这个版本在普通ES6/ES2015中怎么样?

function get(url) {
  return new Promise((resolve, reject) => {
    const req = new XMLHttpRequest();
    req.open('GET', url);
    req.onload = () => req.status === 200 ? resolve(req.response) : reject(Error(req.statusText));
    req.onerror = (e) => reject(Error(`Network Error: ${e}`));
    req.send();
  });
}

函数返回一个promise。下面是一个关于如何使用该函数并处理它返回的承诺的示例:

get('foo.txt')
.then((data) => {
  // Do stuff with data, if foo.txt was successfully loaded.
})
.catch((err) => {
  // Do stuff on error...
});

如果你需要加载一个json文件，你可以使用json .parse()将加载的数据转换为JS对象。

您还可以集成req。responseType='json'到函数中，但不幸的是，没有IE支持它，所以我将坚持使用json .parse()。

2016-08-22 13:59:32

使用“vanilla”(普通)JavaScript:

function loadXMLDoc() {
    var xmlhttp = new XMLHttpRequest();

    xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == XMLHttpRequest.DONE) { // XMLHttpRequest.DONE == 4
           if (xmlhttp.status == 200) {
               document.getElementById("myDiv").innerHTML = xmlhttp.responseText;
           }
           else if (xmlhttp.status == 400) {
              alert('There was an error 400');
           }
           else {
               alert('something else other than 200 was returned');
           }
        }
    };

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

jQuery:

$.ajax({
    url: "test.html",
    context: document.body,
    success: function() {
      $(this).addClass("done");
    }
});

2011-12-19 20:30:43

我怎么能使AJAX调用没有jQuery?

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