Question

我正在使用外部API中的一些REST端点，我正在使用Rest Template接口来实现此目的。当我从这些调用中收到某些HTTP状态代码时，我希望能够抛出自定义应用程序异常。为了实现它，我正在实现ResponseErrorHandler接口，如下所示：

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;
}

我的应用环境：

<?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此自定义错误处理的行为。每个单独的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，它在自定义响应错误处理程序中引发，即：我无法在其余的模板方法中捕获我的自定义异常调用

有关如何捕获这些异常的任何线索？ Spring RestTemplate invoking webservice with errors and analyze status code是高度相关的，但从我的观点来看，虽然它被提议作为解决方案，但我遇到了同样的问题。

[1]：

Answer 1

您已将自定义Exception延伸至IOException

public class MyCustomException extends IOException {

ResponseErrorHandler#handleError()方法是从RestTemplate#handleResponseError(..)调用RestTemplate#doExecute(..)调用的try-catch方法。此根调用包含在IOException块中，该块捕获ResourceAccessException并将其重新包裹在RestClientException中，该RestClientException为cause。

一种可能性是抓住Exception并获取其RuntimeException。

另一种可能性是让您的自定义{{1}}成为{{1}}的子类型。

Answer 2

如果使用springmvc，则可以使用注释@ControllerAdvice创建控制器。在控制器中写：

@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";
}

和DefaultResponseErrorHandler抛出这两种例外：

@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 + "]");
    }
}

您可以使用：e.getResponseBodyAsString();，e.getStatusCode(); blabla在异常发生时获取响应消息。

Rest模板自定义异常处理

2 个答案: