Spring Resttemplate异常处理

下面是代码片段；基本上，当错误代码不是200时，我正在尝试传播异常。

ResponseEntity<Object> response = restTemplate.exchange(url.toString().replace("{version}", version),
                    HttpMethod.POST, entity, Object.class);
            if(response.getStatusCode().value()!= 200){
                logger.debug("Encountered Error while Calling API");
                throw new ApplicationException();
            }

但是，如果来自服务器的响应为500，则会出现异常

org.springframework.web.client.HttpServerErrorException: 500 Internal Server Error
    at org.springframework.web.client.DefaultResponseErrorHandler.handleError(DefaultResponseErrorHandler.java:94) ~[spring-web-4.2.3.RELEASE.jar:4.2.3.RELEASE]

我真的需要在尝试中包装其余模板交换方法吗？那么代码的目的是什么？

您想要创建一个实现的类，ResponseErrorHandler然后使用它的一个实例来设置其余模板的错误处理：

public class MyErrorHandler implements ResponseErrorHandler {
  @Override
  public void handleError(ClientHttpResponse response) throws IOException {
    // your error handling here
  }

  @Override
  public boolean hasError(ClientHttpResponse response) throws IOException {
     ...
  }
}

[...]

public static void main(String args[]) {
  RestTemplate restTemplate = new RestTemplate();
  restTemplate.setErrorHandler(new MyErrorHandler());
}

而且，Spring仅提供了一个类DefaultResponseErrorHandler，您可以扩展该类而不是实现接口，以防万一您只想覆盖该handleError方法。

public class MyErrorHandler extends DefaultResponseErrorHandler {
  @Override
  public void handleError(ClientHttpResponse response) throws IOException {
    // your error handling here
  }
}

查看其源代码，以了解Spring如何处理HTTP错误。

您应该捕获一个HttpStatusCodeException例外：

try {
    restTemplate.exchange(...);
} catch (HttpStatusCodeException exception) {
    int statusCode = exception.getStatusCode().value();
    ...
}

Spring巧妙地将http错误代码视为异常，并假定您的异常处理代码具有处理错误的上下文。要使交换功能按预期运行，请执行以下操作：

    try {
        return restTemplate.exchange(url, httpMethod, httpEntity, String.class);
    } catch(HttpStatusCodeException e) {
        return ResponseEntity.status(e.getRawStatusCode()).headers(e.getResponseHeaders())
                .body(e.getResponseBodyAsString());
    }

这将从响应中返回所有预期结果。

另一种解决方案是本文结尾处“ enlian”描述的解决方案：http ://springinpractice.com/2013/10/07/handling-json-error-object-responses-with-springs-resttemplate

try{
     restTemplate.exchange(...)
} catch(HttpStatusCodeException e){
     String errorpayload = e.getResponseBodyAsString();
     //do whatever you want
} catch(RestClientException e){
     //no response payload, tell the user sth else 
}

Spring从非常大的http状态代码列表中抽象出您。那就是例外的想法。看一下org.springframework.web.client.RestClientException层次结构：

在处理http响应时，有一堆类可以映射最常见的情况。http代码列表确实很大，您不想编写代码来处理每种情况。但是，例如，看看HttpClientErrorException子层次结构。您只有一个例外来映射任何4xx类型的错误。如果您需要深入，那么可以。但是只要捕获HttpClientErrorException，就可以处理将错误数据提供给服务的任何情况。

DefaultResponseErrorHandler非常简单可靠。如果响应状态代码不是2xx系列的，则只对hasError方法返回true。

如果将池（http客户端工厂）或负载平衡（eureka）机制与一起使用RestTemplate，则将无法创建new RestTemplate每个类。如果您要呼叫多个服务，那么您将无法使用该服务，setErrorHandler因为该服务将被全局用于所有请求。

在这种情况下，捕获HttpStatusCodeException似乎是更好的选择。

您唯一的其他选择是RestTemplate使用@Qualifier注释定义多个实例。

另外-但这是我自己的口味-我喜欢我的错误处理紧紧地贴在我的电话上。

我已经如下处理：

try {
  response = restTemplate.postForEntity(requestUrl, new HttpEntity<>(requestBody, headers), String.class);
} catch (HttpStatusCodeException ex) {
  response = new ResponseEntity<String>(ex.getResponseBodyAsString(), ex.getResponseHeaders(), ex.getStatusCode());
}

兑换代码如下：

public <T> ResponseEntity<T> exchange(String url, HttpMethod method,
            HttpEntity<?> requestEntity, Class<T> responseType, Object... uriVariables) throws RestClientException

异常RestClientException具有HttpClientErrorException和HttpStatusCodeException异常。

因此，在RestTemplete有可能occure HttpClientErrorException和HttpStatusCodeException例外。在异常对象中，您可以使用以下方式获取确切的错误消息：exception.getResponseBodyAsString()

这是示例代码：

public Object callToRestService(HttpMethod httpMethod, String url, Object requestObject, Class<?> responseObject) {

        printLog( "Url : " + url);
        printLog( "callToRestService Request : " + new GsonBuilder().setPrettyPrinting().create().toJson(requestObject));

        try {

            RestTemplate restTemplate = new RestTemplate();
            restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());
            restTemplate.getMessageConverters().add(new StringHttpMessageConverter());


            HttpHeaders requestHeaders = new HttpHeaders();
            requestHeaders.setContentType(MediaType.APPLICATION_JSON);

            HttpEntity<Object> entity = new HttpEntity<>(requestObject, requestHeaders);

            long start = System.currentTimeMillis();

            ResponseEntity<?> responseEntity = restTemplate.exchange(url, httpMethod, entity, responseObject);

            printLog( "callToRestService Status : " + responseEntity.getStatusCodeValue());


            printLog( "callToRestService Body : " + new GsonBuilder().setPrettyPrinting().create().toJson(responseEntity.getBody()));

            long elapsedTime = System.currentTimeMillis() - start;
            printLog( "callToRestService Execution time: " + elapsedTime + " Milliseconds)");

            if (responseEntity.getStatusCodeValue() == 200 && responseEntity.getBody() != null) {
                return responseEntity.getBody();
            }

        } catch (HttpClientErrorException exception) {
            printLog( "callToRestService Error :" + exception.getResponseBodyAsString());
            //Handle exception here
        }catch (HttpStatusCodeException exception) {
            printLog( "callToRestService Error :" + exception.getResponseBodyAsString());
            //Handle exception here
        }
        return null;
    }

这是代码说明：

在这种方法中，您必须传递请求和响应类。此方法将自动将响应解析为请求的对象。

首先，您必须添加消息转换器。

restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());
            restTemplate.getMessageConverters().add(new StringHttpMessageConverter());

然后，您必须添加requestHeader。这是代码：

HttpHeaders requestHeaders = new HttpHeaders();
            requestHeaders.setContentType(MediaType.APPLICATION_JSON);

            HttpEntity<Object> entity = new HttpEntity<>(requestObject, requestHeaders);

最后，您必须调用交换方法：

ResponseEntity<?> responseEntity = restTemplate.exchange(url, httpMethod, entity, responseObject);

对于精美的打印，我使用了Gson库。这是gradle：compile 'com.google.code.gson:gson:2.4'

您只需调用以下代码即可获得响应：

ResponseObject response=new RestExample().callToRestService(HttpMethod.POST,"URL_HERE",new RequestObject(),ResponseObject.class);

这是完整的工作代码：

import com.google.gson.GsonBuilder;
import org.springframework.http.*;
import org.springframework.http.converter.StringHttpMessageConverter;
import org.springframework.http.converter.json.MappingJackson2HttpMessageConverter;
import org.springframework.web.client.HttpClientErrorException;
import org.springframework.web.client.HttpStatusCodeException;
import org.springframework.web.client.RestTemplate;


public class RestExample {

    public RestExample() {

    }

    public Object callToRestService(HttpMethod httpMethod, String url, Object requestObject, Class<?> responseObject) {

        printLog( "Url : " + url);
        printLog( "callToRestService Request : " + new GsonBuilder().setPrettyPrinting().create().toJson(requestObject));

        try {

            RestTemplate restTemplate = new RestTemplate();
            restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());
            restTemplate.getMessageConverters().add(new StringHttpMessageConverter());


            HttpHeaders requestHeaders = new HttpHeaders();
            requestHeaders.setContentType(MediaType.APPLICATION_JSON);

            HttpEntity<Object> entity = new HttpEntity<>(requestObject, requestHeaders);

            long start = System.currentTimeMillis();

            ResponseEntity<?> responseEntity = restTemplate.exchange(url, httpMethod, entity, responseObject);

            printLog( "callToRestService Status : " + responseEntity.getStatusCodeValue());


            printLog( "callToRestService Body : " + new GsonBuilder().setPrettyPrinting().create().toJson(responseEntity.getBody()));

            long elapsedTime = System.currentTimeMillis() - start;
            printLog( "callToRestService Execution time: " + elapsedTime + " Milliseconds)");

            if (responseEntity.getStatusCodeValue() == 200 && responseEntity.getBody() != null) {
                return responseEntity.getBody();
            }

        } catch (HttpClientErrorException exception) {
            printLog( "callToRestService Error :" + exception.getResponseBodyAsString());
            //Handle exception here
        }catch (HttpStatusCodeException exception) {
            printLog( "callToRestService Error :" + exception.getResponseBodyAsString());
            //Handle exception here
        }
        return null;
    }

    private void printLog(String message){
        System.out.println(message);
    }
}

谢谢 ：）

一个非常简单的解决方案可以是：

try {
     requestEntity = RequestEntity
     .get(new URI("user String"));
    
    return restTemplate.exchange(requestEntity, String.class);
} catch (RestClientResponseException e) {
        return ResponseEntity.status(e.getRawStatusCode()).body(e.getResponseBodyAsString());
}

这是我的HTTPS POST方法，它为任何类型的错误响应返回响应正文。

public String postHTTPSRequest(String url,String requestJson)
{
    //SSL Context
    CloseableHttpClient httpClient = HttpClients.custom().setSSLHostnameVerifier(new NoopHostnameVerifier()).build();
    HttpComponentsClientHttpRequestFactory requestFactory = new HttpComponentsClientHttpRequestFactory();
    requestFactory.setHttpClient(httpClient);
    //Initiate REST Template
    RestTemplate restTemplate = new RestTemplate(requestFactory);
    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_JSON);
    //Send the Request and get the response.
    HttpEntity<String> entity = new HttpEntity<String>(requestJson,headers);
    ResponseEntity<String> response;
    String stringResponse = "";
    try {
        response = restTemplate.postForEntity(url, entity, String.class);
        stringResponse = response.getBody();
    }
    catch (HttpClientErrorException e)
    {
        stringResponse = e.getResponseBodyAsString();
    }
    return stringResponse;
}