Question

前台用ajax post过来的数据后台要怎样接收？

怎么老是报这种错误的？

Answer 1

导入jQuery包

jQuery(document).ready(function($){
var url = "";
var data = {"param":"我是参数","other":"me too"};
$.post(url,data,function(result){
//.............
});
});

private String param;
private String other;
public void Test(){
PrintWriter out = null;
try{
out = getResponse().getWriter();
if(param.equals("我是参数")){
out.print(other);
}
}catch(Exception e){
//...
}finally{
out.flush();
out.close();
}
}

//getter and setter...

jQuery(document).ready(function($){
var url = "";
var data = {"param":"我是参数","other":"me too"};
$.post(url,data,function(result){
alert(result);
});
});

追问

这是？