使用Spring Boot进行Restful Web Services。
尝试设置一个全局自定义异常处理机制,该机制依赖于@RestControllerAdvice,它可以处理已知但未知的异常。
的pom.xml
<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>1.5.4.RELEASE</version>
</parent>
<properties>
<java.version>1.8</java.version>
</properties>
<repositories>
<repository>
<id>spring-releases</id>
<url>https://repo.spring.io/libs-release</url>
</repository>
</repositories>
<pluginRepositories>
<pluginRepository>
<id>spring-releases</id>
<url>https://repo.spring.io/libs-release</url>
</pluginRepository>
</pluginRepositories>
<dependencies>
<!-- Spring -->
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-web</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-test</artifactId>
<scope>test</scope>
</dependency>
</dependencies>
GlobalControllerExceptionHandler:
@RestControllerAdvice
public class GlobalControllerExceptionHandler {
private static final Logger LOG = Logger.getLogger(GlobalControllerExceptionHandler.class);
@ExceptionHandler(value = { ConstraintViolationException.class })
@ResponseStatus(HttpStatus.BAD_REQUEST)
public ApiErrorResponse constraintViolationException(ConstraintViolationException ex) {
LOG.error(ex.getCause().toString());
return new ApiErrorResponse(400, "Bad Request");
}
@ExceptionHandler(value = { NoHandlerFoundException.class })
@ResponseStatus(HttpStatus.NOT_FOUND)
public ApiErrorResponse noHandlerFoundException(Exception ex) {
LOG.error(ex.getCause().toString());
return new ApiErrorResponse(404, "Resource Not Found");
}
@ExceptionHandler(value = { Exception.class })
@ResponseStatus(HttpStatus.INTERNAL_SERVER_ERROR)
public ApiErrorResponse unknownException(Exception ex) {
LOG.error(ex.getCause().toString());
return new ApiErrorResponse(500, "Internal Server Error");
}
}
ApiErrorResponse:
public class ApiErrorResponse {
private int status;
private String message;
public ApiErrorResponse(int status, String message) {
this.status = status;
this.message = message;
}
public int getStatus() {
return status;
}
public String getMessage() {
return message;
}
@Override
public String toString() {
return new ToStringBuilder(this).append(status)
.append(message)
.toString();
}
}
这个问题是当我使用第三方库做某事时, 未知异常可能是404但是以500返回!
e.g。使用具有未知索引的ElasticSearch(故意查看异常类型):
{
"timestamp": 1501236796640,
"status": 500,
"error": "Internal Server Error",
"exception": "org.elasticsearch.client.ResponseException",
"message": "POST http://localhost:9200/fn3r4343/_search?pretty=true: HTTP/1.1 404 Not Found"
{
"error": {
"root_cause": [
{
"type": "index_not_found_exception",
"reason": "no such index",
"resource.type": "index_or_alias",
"resource.id": "fn3r4343",
"index_uuid": "_na_",
"index": "fn3r4343"
}
],
"type": "index_not_found_exception",
"reason": "nosuchindex",
"resource.type": "index_or_alias",
"resource.id": "fn3r4343",
"index_uuid": "_na_",
"index": "fn3r4343"
}
{ "root_cause" :
[
{
"type" :"index_not_found_exception",
"reason" : no such index", "resource.type" : "index_or_alias",
"resource.id" : "fn3r4343",
"index_uuid" : "_na_",
"index" : "fn3r4343"
}
],
[
{
"type" : "index_not_found_exception",
"reason" : "no such index",
"resource.type" : "index_or_alias",
"resource.id" : "fn3r4343",
"index_uuid" : "_na_",
"index" : "fn3r4343"
},
"status": 404
]
}
"path": "/myapp/search"
}
可以看出,这会返回HTTP 500状态,但在有效负载中却是HTTP 404!
我要求的是返回这个:
{
"message" : "Index Not Found Exception",
"status" : "HTTP 404"
}
对于已知的HTTP 404例外:
{
"message" : "Not Found",
"status" : "HTTP 404"
}
使用RestControllerAdvice来捕获任何类型的Exception并将响应自定义为使用REST API对客户端可读/有用的JSON格式是否有良好的做法/机制?
这篇文章并非真正针对弹性搜索,而是试图通过尝试使用@RestControllerAdvice来处理任何类型的Exception,为客户端应用程序提供正确的响应......
答案 0 :(得分:2)
引发的基础异常是org.elasticsearch.client.ResponseException
(您可能正在使用低级REST客户端)。
因此,在您的建议中,您需要为该异常添加处理程序并返回基础状态代码:
@ExceptionHandler(value = { ResponseException.class })
public ApiErrorResponse noHandlerFoundException(Exception ex) {
LOG.error(ex.getCause().toString());
int status = ((ResponseException) ex).getResponse().getStatusLine().getStatusCode();
return new ApiErrorResponse(status, "<some message depending on status code>");
}
答案 1 :(得分:0)
我认为下面的课程应该有所帮助:
@ControllerAdvice
public class CustomExceptionHandler extends ResponseEntityExceptionHandler {
@ExceptionHandler(value = {MyException.class})
protected ResponseEntity<Object> handleConflict(Exception exception, WebRequest request)
{
FinalResponse response = new FinalResponse()//Define this Bean in a way that it contains all required paramteres to be sent in response when exception occurs
response.setErrorCd("123");
response.setMessage("Exception while processing.");
if (exception instanceof MyException)
{
return new ResponseEntity<>(response, HttpStatus.BAD_REQUEST);
}
else
{
return new ResponseEntity<>(response, HttpStatus.INTERNAL_SERVER_ERROR);
}
}
}