ajax设置超时时间，超时返回弹出“错误”

Question

ajax设置超时时间，如果正常就返回正常的回调数据，如果超时返回弹出“错误”，

Answer 1

http://www.cnblogs.com/snandy/archive/2011/04/26/2026734.html
去看看这个，或许对你有帮助

Answer 2

<script type=”text/javascript”>
 
       function getXMLHttpRequest()
       {
            var xhr = false;
            if ( window.XMLHttpRequest )
            {
                xhr = new XMLHttpRequest();
                if ( xhr.overrideMimeType )
                {
                    xhr.overrideMimeType(”text/xml”);
                }
            }else if ( window.ActiveXObject )
            {
                try{
                        xhr = new ActiveXObject(”Msxml2.XMLHTTP”);
                    }catch(e)
                    {
                        try{
                                xhr = new ActiceXObject(”Microsoft.XMLHTTP”);
                            }catch(e)
                            {
                                xhr = false;
                            }
                    }
            }
 
            return xhr;
       }
 
        window.onload = function()
        {
            var _x = getXMLHttpRequest(); //获得XMLHttpRequest对象
            if ( !_x ) return false;
            _x.open(”get”,http://www.baidu.com,true);
            _x.onreadystatechange = function()
            {
                if ( _x.readyState == 4 && _x.status == 200 )
                {
                    clearTimeout(clearTO); //如果准备状态成功，并且HTTP状态码正确则清除setTimeout
                    alert(”Success”);
                }    
            };
            var cleaeTO = setTimeout(function()  /*重点，在请求发布后开始设置setTimeout，如果请求状态不成功也就是readyState != 4 那么setTimeout将会在5秒后运行，并弹出信息提示，要是请求成功，将会清除该setTimeout*/
            {
                _x.abort(); //终止XMLHttpRequest对象
                alert(”系统异常，请您刷新页面或稍后再试….”);
            },50000);
            _x.send(null);
        };
</script>