Spring MVC返回JSONS和异常处理 [英] Spring MVC returning JSONS and exception Handling

问题描述

要进一步阐述，我有一个名为UserDetails的对象，它有两个字段，名为name，emailAddress

  @ResponseBody UserDetails 
  

现在，json返回的样子看起来像

{name：TheUsersName，
emailAddress：[e-mail]}

有什么办法可以在返回之前修改json（所有控制器中的所有方法中的所有jsons），其中将添加status字段，其他json数据将在data json中的关键。

另外，当某个地方的java服务器引发异常时，如何将json返回到前端，json应该具有status：false和异常名称（至少是状态

解决方案

是的。返回一个模型和一个视图。

  public ModelMap getUserDetails（）{
 UserDetails userDetails; //从某个地方获取此对象
 ModelMap map = new ModelMap（）（; 
 map.addAttribute（data，userDetails）; 
 map.addAttribute（success，true）; 
返回地图; 
} 
  

添加你会做的例外与key相同的方式，成功= false。

I am using Spring MVC with Controllers, my question is how do I return a JSON response which is different from the @ResponseBody object which is returned and convereted to a JSON to be returned.

To elaborate further, I have the object called "UserDetails" which has two fields called "name", "emailAddress"

@ResponseBody UserDetails

now the json returned will look like

{ name : "TheUsersName", emailAddress:"[e-mail]" }

Is there any way I can modify the json before returning (ALL jsons in all methods across all controllers) where a "status" field will be added and the other json data will be under the "data" key in the json.

Also how do I return a json to the frontend when the java server from somewhere throws an exception, the json should have "status : false" and the exception name (atleast the status part though)

解决方案

Yes. Return a model and a view instead.

public ModelMap getUserDetails() {
    UserDetails userDetails; // get this object from somewhere
    ModelMap map = new ModelMap()(;
    map.addAttribute("data", userDetails);
    map.addAttribute("success", true);
    return map;
}

To add the exception you'd do it the same way with a key and success = false.

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