如何使用JavaScript进行AJAX调用,而不使用jQuery?


当前回答

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);
     }
 });

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();

尝试使用Fetch Api (Fetch Api)

fetch('http://example.com/movies.json').then(response => response.json()).then(data => console.log(data));

非常清澈,100%香草味。

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);
}

这只是一个简单的4步过程,

我希望这对你们有帮助

步骤1。存储对XMLHttpRequest对象的引用

var xmlHttp = createXmlHttpRequestObject();

步骤2。检索XMLHttpRequest对象

function createXmlHttpRequestObject() {
    // will store the reference to the XMLHttpRequest object
    var xmlHttp;
    // if running Internet Explorer
    if (window.ActiveXObject) {
        try {
            xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
        } catch (e) {
            xmlHttp = false;
        }
    }
    // if running Mozilla or other browsers
    else {
        try {
            xmlHttp = new XMLHttpRequest();
        } catch (e) {
            xmlHttp = false;
        }
    }
    // return the created object or display an error message
    if (!xmlHttp)
        alert("Error creating the XMLHttpRequest object.");
    else
        return xmlHttp;
}

步骤3。使用XMLHttpRequest对象进行异步HTTP请求

function process() {
    // proceed only if the xmlHttp object isn't busy
    if (xmlHttp.readyState == 4 || xmlHttp.readyState == 0) {
        // retrieve the name typed by the user on the form
        item = encodeURIComponent(document.getElementById("input_item").value);
        // execute the your_file.php page from the server
        xmlHttp.open("GET", "your_file.php?item=" + item, true);
        // define the method to handle server responses
        xmlHttp.onreadystatechange = handleServerResponse;
        // make the server request
        xmlHttp.send(null);
    }
}

步骤4。当从服务器接收消息时自动执行

function handleServerResponse() {

    // move forward only if the transaction has completed
    if (xmlHttp.readyState == 4) {
        // status of 200 indicates the transaction completed successfully
        if (xmlHttp.status == 200) {
            // extract the XML retrieved from the server
            xmlResponse = xmlHttp.responseText;
            document.getElementById("put_response").innerHTML = xmlResponse;
            // restart sequence
        }
        // a HTTP status different than 200 signals an error
        else {
            alert("There was a problem accessing the server: " + xmlHttp.statusText);
        }
    }
}