解决ajax请求后台,有时收不到返回值的问题

$.ajax({ 
  type:"POST", 
  url:"Venue.aspx", 
  dataType:"html", 
  success:function(result){ //function1()
  f1(); 
  f2(); 
 } 
  failure:function (result) { 
  alert('Failed'); 
  }, 
 } 
 function2();

$.ajax({ 
  type:"POST", 
  url:"Venue.aspx", 
  dataType:"html", 
  async: false,
 success:function(result){ //function1()
  f1(); 
  f2(); 
  } 
 failure:function (result) { 
  alert('Failed'); 
  }, 
 } 
 function2(); 

var returnValue = null; 
xmlhttp = createXmlHttp(); 
xmlhttp.onreadystatechange = function() { 
 if(xmlhttp.readyState == 4 && xmlhttp.status == 200) { 
 if (xmlhttp.responseText == "true") { 
  returnValue = "true"; 
 } 
 else { 
  returnValue = "false"; 
 } 
 } 
}; 
xmlhttp.open("Post",url,true); //异步传输 
xmlhttp.setRequestHeader("If-Modified-Since","0"); //不缓存Ajax
xmlhttp.send(sendStr); 
return returnValue;
在异步时才可以用xmlHttpReq.onreadystatechange状态值！下面是异步和同步的不同调用方式：

xmlHttpReq.open("GET",url,true);//异步方式
 xmlHttpReq.onreadystatechange = showResult; //showResult是回调函数名
 xmlHttpReq.send(null);
function showResult(){ 
 if(xmlHttpReq.readyState == 4){ 
 if(xmlHttpReq.status == 200){
 ******
 }
 }
}

xmlHttpReq.open("GET",url,false);//同步方式 
  xmlHttpReq.send(null); 
  showResult(); //showResult虽然是回调函数名但是具体用法不一样~ 
function showResult(){ 
  //if(xmlHttpReq.readyState == 4){ 这里就不用了，直接dosomething吧~ 
  //if(xmlHttpReq.status == 200){ 
   ******//dosomething 
  //} 
  //} 
}
xmlhttp.open("Post",url,true);