AJAX

XMLHttpRequest发送请求

• open(method,url,async)

• send(string)

代码演示！

request.open("GET","get.php",true);
request.send();  

request.open("POST","post.php",true);
request.send();

requset.open("POST","create.php",true);
request.setRequestHeader("Content-type","application/x-www-form-rulencoded");
request.send("name=王二狗&sex=男");