在春季启动时发送响应的最佳做法

问题描述 投票:0回答:2

我在Spring boot中编写REST Api-s。我想确保使用swagger API开发工具(Swagger)让前端开发人员可以读取我的代码。例如

@GetMapping("/getOne")
    public ResponseEntity<?> getOne(@RequestParam String id) {
        try {
            return new ResponseEntity<Branch>(branchService.getOne(id), HttpStatus.OK);
        } catch (Exception e) {
            return new ResponseEntity<FindError>(new FindError(e.getMessage()), HttpStatus.BAD_REQUEST);
        }
    }

如果请求成功,则response是Branch对象,如果失败,则响应是FindError对象,该对象只有一个属性(消息)。所以两者都可以进行取决于响应。但是,虚张声势的UI并没有显示应该如何显示响应,因为我使用“?”作为通用类型。这是捕获错误的最佳做法吗? (这种编码文档swagger对前端开发人员没用,因为它没有显示响应对象)。或者针对上述问题的任何最佳做法?

有许多返回不同对象的方法,如Branch。提前致谢

spring spring-boot swagger swagger-ui swagger-2.0
2个回答
3
投票

首先,您应该遵循RESTful API的最佳实践。不要使用动词,而是使用名词作为URL.So而不是@GetMapping("/getOne"),你可以把它写成@GetMapping("/branch/{id}")。你可以参考这个博客https://blog.mwaysolutions.com/2014/06/05/10-best-practices-for-better-restful-api/

@ 2ndly,不要返回通用类型吗? ,您可以使用特定类型,此处为分支并执行中央异常处理。以下代码段可以帮助您:

@GetMapping("/branch/{id}")
public ResponseEntity<Branch> getBranch(@Pathvariable String id) {
{
    Branch branch = branchService.getOne(id);

    if(branch == null) {
         throw new RecordNotFoundException("Invalid Branch id : " + id);
    }
    return new ResponseEntity<Branch>(branch, HttpStatus.OK);
}

record not found exception.Java

@ResponseStatus(HttpStatus.NOT_FOUND)
public class RecordNotFoundException extends RuntimeException
{
    public RecordNotFoundException(String exception) {
        super(exception);
    }
}

custom exception handler.Java

@ControllerAdvice
public class CustomExceptionHandler extends ResponseEntityExceptionHandler
{
    @ExceptionHandler(Exception.class)
    public final ResponseEntity<Object> handleAllExceptions(Exception ex, WebRequest request) {
        List<String> details = new ArrayList<>();
        details.add(ex.getLocalizedMessage());
        ErrorResponse error = new ErrorResponse("Server Error", details);
        return new ResponseEntity(error, HttpStatus.INTERNAL_SERVER_ERROR);
    }

    @ExceptionHandler(RecordNotFoundException.class)
    public final ResponseEntity<Object> handleRecordNotFoundException(RecordNotFoundException ex, WebRequest request) {
        List<String> details = new ArrayList<>();
        details.add(ex.getLocalizedMessage());
        ErrorResponse error = new ErrorResponse("Record Not Found", details);
        return new ResponseEntity(error, HttpStatus.NOT_FOUND);
    }
}

error response.Java

public class ErrorResponse
{
    public ErrorResponse(String message, List<String> details) {
        super();
        this.message = message;
        this.details = details;
    }

    private String message;

    private List<String> details;

    //Getter and setters
}

上面的类处理多个异常,包括RecordNotFoundException,您也可以自定义有效负载验证。

测试用例 :

1) HTTP GET /branch/1 [VALID]

HTTP Status : 200

{
    "id": 1,
    "name": "Branch 1",
    ...
}
2) HTTP GET /branch/23 [INVALID]

HTTP Status : 404

{
    "message": "Record Not Found",
    "details": [
        "Invalid Branch id : 23"
    ]
}

0
投票

我建议这样做。

@GetMapping("/getOne")
public Response getOne(@RequestParam String id) {
        ResponseEntity<Branch> resbranch;
        ResponseEntity<FindError> reserror;
        try {
            resbranch=new ResponseEntity<Branch>(branchService.getOne(id), HttpStatus.OK);
            return Response.status(200).entity(resbranch).build();

        } catch (Exception e) {
            reserror=new ResponseEntity<FindError>(new FindError(e.getMessage()), HttpStatus.BAD_REQUEST);
            return Response.status(400).entity(reserror).build();
        }
    }

200表示OK,400表示错误请求。这里不再是模棱两可的类型..

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