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

<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

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

xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
        alert(this.responseText);
    }
};
xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();

2011-12-19 20:30:30

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

这可能会有帮助:

function doAjax(url, callback) {
    var xmlhttp = window.XMLHttpRequest ? new XMLHttpRequest() : new ActiveXObject("Microsoft.XMLHTTP");

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

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

2013-11-29 18:49:16

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();

2015-05-18 01:18:07

我怎么能使AJAX调用没有jQuery?

推荐文章

最新文章

标签