

@RequestMapping(value = "/SubmitStep1.json", method = RequestMethod.POST, headers = "Accept=application/json,application/xml")
@ResponseBody
public List<ShopDetails> showShopList(@RequestBody ShopDetails shopDetails)throws Exception{
List<ShopDetails> shopDetailsList=new ArrayList<ShopDetails>();
shopDetailsList=dbq.getShopDetails(shopDetails);
return shopDetailsList;
}在上面的代码中,我返回了店铺列表,其中包含每个店铺的详细信息。
所以,我的问题是,如果我得到商店列表,我是否可以在返回时添加成功/错误消息。
发布于 2017-08-21 23:21:05
正如@araknoid所说的--你可以创建包装器:
public class ShopListResponse {
private List<ShopDetails> shopList;
private String message;
public ShopListResponse (List<ShopDetails> shopList, String message){
this.shopList = shopList;
this.message = message;
}
// getters and setters
}在您的控制器类中:
@RequestMapping(value = "/SubmitStep1.json", method = RequestMethod.POST, headers = "Accept=application/json,application/xml")
@ResponseBody
public ResponseEntity<ShopListResponse> showShopList(@RequestBody ShopDetails shopDetails)throws Exception{
List<ShopDetails> shopDetailsList = dbq.getShopDetails(shopDetails);
return new ResponseEntity<>(new ShopListResponse(shopDetailsList, "Success or error message"), HttpStatus.OK);
}如果你想返回一个错误--你可以返回HttpStatus.NOT_FOUND,或者只是返回带有错误的HttpStatus.OK和消息--这取决于你的方法。
这里是控制器的结果:

https://stackoverflow.com/questions/45799507
复制相似问题