如何使用JavaScript进行AJAX调用,而不使用jQuery?


当前回答

你可以根据浏览器获得正确的对象

function getXmlDoc() {
  var xmlDoc;

  if (window.XMLHttpRequest) {
    // code for IE7+, Firefox, Chrome, Opera, Safari
    xmlDoc = new XMLHttpRequest();
  }
  else {
    // code for IE6, IE5
    xmlDoc = new ActiveXObject("Microsoft.XMLHTTP");
  }

  return xmlDoc;
}

有了正确的对象,GET可以被抽象为:

function myGet(url, callback) {
  var xmlDoc = getXmlDoc();

  xmlDoc.open('GET', url, true);

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

  xmlDoc.send();
}

并将邮件发送到:

function myPost(url, data, callback) {
  var xmlDoc = getXmlDoc();

  xmlDoc.open('POST', url, true);
  xmlDoc.setRequestHeader("Content-type", "application/x-www-form-urlencoded");

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

  xmlDoc.send(data);
}

其他回答

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)
};

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()。

在浏览器中使用纯JavaScript:

var xhr = new XMLHttpRequest();

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

xhr.open("GET", "mock_data.json", true);

xhr.send();

或者如果你想使用Browserify使用node.js来捆绑你的模块。你可以使用超级代理:

var request = require('superagent');
var url = '/mock_data.json';

 request
   .get(url)
   .end(function(err, res){
     if (res.ok) {
       console.log('yay got ' + JSON.stringify(res.body));
     } else {
       console.log('Oh no! error ' + res.text);
     }
 });
xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
        alert(this.responseText);
    }
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();

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));