Ajax 实现?

 1 var xhr = new XMLHttpRequest();
 2 xhr.onreadystatechange = function(){
 3   if(xhr.readyState == 4){
 4     if((xhr.status >= 200 && xhr.status <= 300) || xhr.status == 304){
 5       alert(xhr.responseText);
 6     }else{
 7       alert("Request was unsuccessful:" + xhr.status);
 8     }
 9   }
10   //get
11   xhr.open("get", "example.php", true);
12   xhr.setRequestHeader("MyHeader", "MyValue");
13   xhr.send(null);
14   //post
15   // xhr.open("post", "postexample.php", true);
16   // xhr.setRequestHeader("Content-Type", "applicatoin/x-www-form-urlencoded");
17   // var form = document.getElementById("user-info");
18   // xhr.send(serialize(form));
19 };

 

posted @ 2015-11-30 10:40  将军喊俺哥  阅读(124)  评论(0编辑  收藏  举报