如何在Java中重新抛出异常

Question

    responses:
        '200':
            content:
                application/json:
                    schema:
                        $ref: '#/components/schemas/MyResponse'
            description: OK
        '401':
            content:
                application/json:
                    schema:
                        $ref: '#/components/schemas/MyError'

    @ApiResponses(value = { 
        @ApiResponse(responseCode = "200", description = "Authentication succeeded", content = @Content(mediaType = "application/json", schema = @Schema(implementation = MyResponse.class))),
        
        @ApiResponse(responseCode = "401", description = "Authentication failed", content = @Content(mediaType = "application/json", schema = @Schema(implementation = MyError.class))) })
    @RequestMapping(value = "/login", method = RequestMethod.POST)
    default ResponseEntity<MyResponse> LoginMethod(//some parameters...) { //something}

    public ResponseEntity<MyResponse> LoginMethod(//some parameters...) {
try { 
  //call external API which throw an exception 
} catch(ApiException e){
  e.getResponseBody; // This is a string type of MyError class in JSON format returned
  // throw e;
}

@apiResponse似乎也没有更正响应类型。

如本问题所述，如何在swagger Codegen中处理多个响应/返回类型（204为空，400为非空等）？

我可以这样扔

throw new ResponseStatusException(HttpStatus.valueOf(e.getCode()), e.getResponseBody());

但是有没有更好的方法来做到这一点呢？我只想将.getResponseBody（）作为对象而不是字符串返回。

非常感谢。

Answer 1

您可以在throws声明中添加apiexception，如下所示：

public ResponseEntity<MyResponse> LoginMethod(//some parameters...) throws ApiException {
   // here your code that can create teh ApiException
}

现在调用这个的方法也会请求throw异常。您将能够在其中管理异常。

您还可以创建一个包含所有所需信息的新对象。它还将信息格式化为始终相同，而不依赖于抛出的错误。