我的 Spring REST 服務(wù)的當(dāng)前響應(yīng)如下：[ { "id": "5cc81d256aaed62f8e6462f4", "email": "exmaplefdd@gmail.com" }, { "id": "5cc81d386aaed62f8e6462f5", "email": "exmaplefdd@gmail.com" }]我想將它包裝在一個(gè) json 對(duì)象中，如下所示： { "elements":[ { "id": "5cc81d256aaed62f8e6462f4", "email": "exmaplefdd@gmail.com" }, { "id": "5cc81d386aaed62f8e6462f5", "email": "exmaplefdd@gmail.com" } ]} 控制器： @RequestMapping(value = "/users", method = GET,produces = "application/xml") @ResponseBody public ResponseEntity<List<User>> getPartnersByDate(@RequestParam("type") String type, @RequestParam("id") String id) throws ParseException { List<User> usersList = userService.getUsersByType(type); return new ResponseEntity<List<User>>(usersList, HttpStatus.OK);}用戶模型類：@Document(collection = "user")public class User { @Id private String id; private String email;}我該如何實(shí)施？
1 回答
幕布斯7119047
TA貢獻(xiàn)1794條經(jīng)驗(yàn) 獲得超8個(gè)贊
您可以創(chuàng)建一個(gè)新的對(duì)象來序列化：
class ResponseWrapper {
private List<User> elements;
ResponseWrapper(List<User> elements) {
this.elements = elements;
}
}
ResponseWrapper然后在你的控制器方法中返回一個(gè)實(shí)例：
@RequestMapping(value = "/users", method = GET,produces = "application/xml")
@ResponseBody
public ResponseEntity<ResponseWrapper> getPartnersByDate(@RequestParam("type") String type, @RequestParam("id") String id) throws ParseException {
List<User> usersList = userService.getUsersByType(type);
ResponseWrapper wrapper = new ResponseWrapper(usersList);
return new ResponseEntity<ResponseWrapper>(wrapper, HttpStatus.OK);
}
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