spring接受複雜json引數,轉化為物件
阿新 • • 發佈:2019-02-20
我們在springmvc要接受一個複雜的引數,但是不知道怎麼接受這個複雜的引數,那我們可以返回這個引數,這樣就知道怎麼寫請求引數
例如我要一個json字串引數,引數是一個物件,一個物件包含2個物件,包含一個user和localauth物件
@ResponseBody @RequestMapping(value="/registerOrdinaryUser",method = RequestMethod.GET) public UserVo registerOrdinaryUser(){ UserVo userVo = new UserVo(); Localauth localauth = new Localauth(); localauth.setUsername("jacket"); User user = new User(); user.setSex("男"); userVo.setLocalauth(localauth); userVo.setUser(user); return userVo; }
返回json
{ "localauth": { "id": null, "userId": null, "username": "jacket", "email": null, "password": null }, "user": { "userId": null, "status": null, "name": null, "sex": "男", "role": null, "birthday": null, "regdate": null, "phone": null, "head": null } }
這樣就知道請求引數怎麼寫了。
下面測試一下。
@ResponseBody
@RequestMapping(value="/registerOrdinaryUser",method = RequestMethod.GET)
public UserVo registerOrdinaryUser(@RequestBody UserVo userVo){
return userVo;
}
傳遞引數,確認返回的引數是否接受,
請求引數
返回引數{ "localauth": { "username": "struts", "password": 123, "email": "" }, "user": { "sex": "男" } }
{
"localauth": {
"id": null,
"userId": null,
"username": "struts",
"email": "",
"password": "123"
},
"user": {
"userId": null,
"status": null,
"name": null,
"sex": "男",
"role": null,
"birthday": null,
"regdate": null,
"phone": null,
"head": null
}
}