Rest模板自定义异常处理
Rest模板自定义异常处理
提问于 2014-01-29 19:35:38
我正在使用来自外部API的一些REST端点，并且我正在使用Rest模板接口来实现此目的。当我从这些调用中接收到某些HTTP状态码时，我希望能够抛出自定义应用程序异常。为了实现它，我按如下方式实现了ResponseErrorHandler接口：
public class MyCustomResponseErrorHandler implements ResponseErrorHandler {
private ResponseErrorHandler myErrorHandler = new DefaultResponseErrorHandler();
public boolean hasError(ClientHttpResponse response) throws IOException {
return myErrorHandler.hasError(response);
}
public void handleError(ClientHttpResponse response) throws IOException {
String body = IOUtils.toString(response.getBody());
MyCustomException exception = new MyCustomException(response.getStatusCode(), body, body);
throw exception;
}
}
public class MyCustomException extends IOException {
private HttpStatus statusCode;
private String body;
public MyCustomException(String msg) {
super(msg);
// TODO Auto-generated constructor stub
}
public MyCustomException(HttpStatus statusCode, String body, String msg) {
super(msg);
this.statusCode = statusCode;
this.body = body;
}
public HttpStatus getStatusCode() {
return statusCode;
}
public void setStatusCode(HttpStatus statusCode) {
this.statusCode = statusCode;
}
public String getBody() {
return body;
}
public void setBody(String body) {
this.body = body;
}
}最后，这是客户端代码(省略了不相关的代码)：
public LoginResponse doLogin(String email, String password) {
HttpEntity<?> requestEntity = new HttpEntity<Object>(crateBasicAuthHeaders(email,password));
try{
ResponseEntity<LoginResponse> responseEntity = restTemplate.exchange(myBaseURL + "/user/account/" + email, HttpMethod.GET, requestEntity, LoginResponse.class);
return responseEntity.getBody();
} catch (Exception e) {
//Custom error handler in action, here we're supposed to receive a MyCustomException
if (e instanceof MyCustomException){
MyCustomException exception = (MyCustomException) e;
logger.info("An error occurred while calling api/user/account API endpoint: " + e.getMessage());
} else {
logger.info("An error occurred while trying to parse Login Response JSON object");
}
}
return null;
}我的应用上下文：
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:context="http://www.springframework.org/schema/context"
xmlns:mvc="http://www.springframework.org/schema/mvc" xmlns:spring="http://camel.apache.org/schema/spring"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.2.xsd
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.2.xsd">
<!-- Rest template (used in bridge communication) -->
<bean id="restTemplate" class="org.springframework.web.client.RestTemplate">
<property name="errorHandler" ref="myCustomResponseErrorHandler"></property>
</bean>
<!-- Bridge service -->
<bean id="myBridgeService" class="a.b.c.d.service.impl.MyBridgeServiceImpl"/>
<!-- Bridge error handler -->
<bean id="myCustomResponseErrorHandler" class="a.b.c.d.service.handlers.MyCustomResponseErrorHandler"/>
</beans>我怀疑我没有正确理解这个自定义错误处理的行为。每个rest模板方法都可能抛出一个RestClientException，它遵循异常层次结构，是RuntimeException的子类，而不是IOException，后者在自定义响应错误处理程序中抛出，即:我无法在rest模板方法调用中捕获自定义异常。
有什么关于如何捕获这些异常的线索吗？Spring RestTemplate invoking webservice with errors and analyze status code是高度相关的，但从我的角度来看，它经历了同样的问题，尽管它是作为解决方案提出的。
1
回答 2
Stack Overflow用户
回答已采纳
发布于 2014-01-29 21:44:39
您已经从IOException扩展了您的自定义Exception
public class MyCustomException extends IOException {ResponseErrorHandler#handleError()方法是从RestTemplate#handleResponseError(..)调用的，后者由RestTemplate#doExecute(..)调用。这个根调用被包装在一个try-catch块中，该块捕获IOException并将其重新抛出，包装在一个ResourceAccessException中，这是一个RestClientException。
一种可能是捕获RestClientException并获取其cause。
另一种可能是使您的自定义Exception成为RuntimeException的子类型。
Stack Overflow用户
发布于 2016-09-22 20:01:35
如果使用springmvc，则可以使用注释@ControllerAdvice创建控制器。在控制器中写入：
@ExceptionHandler(HttpClientErrorException.class)
public String handleXXException(HttpClientErrorException e) {
log.error("log HttpClientErrorException: ", e);
return "HttpClientErrorException_message";
}
@ExceptionHandler(HttpServerErrorException.class)
public String handleXXException(HttpServerErrorException e) {
log.error("log HttpServerErrorException: ", e);
return "HttpServerErrorException_message";
}
...
// catch unknown error
@ExceptionHandler(Exception.class)
public String handleException(Exception e) {
log.error("log unknown error", e);
return "unknown_error_message";
}DefaultResponseErrorHandler抛出以下两种异常：
@Override
public void handleError(ClientHttpResponse response) throws IOException {
HttpStatus statusCode = getHttpStatusCode(response);
switch (statusCode.series()) {
case CLIENT_ERROR:
throw new HttpClientErrorException(statusCode, response.getStatusText(),
response.getHeaders(), getResponseBody(response), getCharset(response));
case SERVER_ERROR:
throw new HttpServerErrorException(statusCode, response.getStatusText(),
response.getHeaders(), getResponseBody(response), getCharset(response));
default:
throw new RestClientException("Unknown status code [" + statusCode + "]");
}
}当异常发生时，你可以使用：e.getResponseBodyAsString();，e.getStatusCode(); blabla来获取响应消息。
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原文链接：
https://stackoverflow.com/questions/21429899
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