如何在程序中申明AJAX

<script type="text/javascript">
var ss="";
        var xmlHttp;
           function createXMLHttpRequest() {
            if (window.ActiveXObject) {
                xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
            }
            else if (window.XMLHttpRequest) {
                xmlHttp = new XMLHttpRequest();               
            }
        }

        function go() {
            createXMLHttpRequest();
            var url = "../ReturnXY";
            xmlHttp.open("GET", url, true);
          
            xmlHttp.onreadystatechange = goCallback;
            xmlHttp.send(null);
        }

        function goCallback() {
       
            if (xmlHttp.readyState == 4) {
            alert(xmlHttp.status);
                if (xmlHttp.status == 200) {
                    ss=xmlHttp.responseText;
                }
            }
        }
       
    function dd()
    {
    alert(ss);
    }
    </script>

评论

1 楼 IvanLi 2006-10-13  

用一下prototype.js岂不是更好，何必重复发明轮子？