jquery ajax封装问题

Question

ajax封装

Answer 1

function ajaxtest(url){
  return $.ajax({ 
           type: "get",
            url:url,
            timeout: 5000,
            data: {type:"info"}, 
            cache:false,
            success:function(xmlDoc, textStatus, xhr){ 
                if(xhr.readyState==4)
                {
                    if(xhr.status == 200)
                    {
                      var str=xhr.responseText; //获取返回值
                    }//200
                }//4
            }, 
            error: function(xhr, textStatus, errorThrown)
            {
                if("timeout" == textStatus)
                {
                    alert("请求超时");
                }
                else
                {
                    alert("网络异常");
                }
            }
        }); //ajax结束
}

Answer 2

function ajaxtest(url,data,callback){
  $.ajax({ 
           type: "get",
            url:url,
            timeout: 5000,
            data: data, 
            cache:false,
            success:function(xmlDoc, textStatus, xhr){ 
                if(xhr.readyState==4)
                {
                    if(xhr.status == 200)
                    {
                      var str=xhr.responseText; //获取返回值
                      if(callback){
                             callback(str);
                      }
                    }//200
                }//4
            }, 
            error: function(xhr, textStatus, errorThrown)
            {
                if("timeout" == textStatus)
                {
                    alert("请求超时");
                }
                else
                {
                    alert("网络异常");
                }
            }
        }); //ajax结束
}

Answer 3

jquery ajax已经封装的够通用了$.ajax
$.post
$.get
$.getJSON
$.getScript都可以用， 没有必要自己再去封闭一遍