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

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);
}

2013-09-25 02:26:14

其他回答

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

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);
     }
 });

2015-04-27 05:18:38

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

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);
}

2013-09-25 02:26:14

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

我怎么能使AJAX调用没有jQuery?

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