Question

使用Spring Rest Template调用Rest Web服务，如下所示 -

ResponseEntity<String> response = restTemplate.exchange(builder.build().encode().toUri(), HttpMethod.GET, entity, String.class);

并以String格式获取输出

<Info xmlns="http://schemas.test.org/2009/09/Tests.new" xmlns:i="http://www.w3.org/2001/XMLSchema-instance">
<FirstName>FirstName</FirstName>
<LastName>LastName</LastName>
<TestGuid>Guid</TestGuid>
<TestUID>1</TestUID>
<Token>token</Token>
<TestUserID>14</TestUserID>
</Info>

尝试将其解组为java类时，如下所示

ResponseEntity<Info> response = restTemplate.exchange(builder.build().encode().toUri(), HttpMethod.GET, entity, Info.class)

Info类定义为

@XmlRootElement(name = "Info", namespace = "http://schemas.test.org/2009/09/Tests.new")
public class Info implements Serializable{

    private String FirstName;
    private String LastName;
    private String TestGuid;
    private String TestUID;
    private String Token;
    private String TestUserID;

    //getters and setter

}

如果Response http代码为500，则响应不是info类型，而是其他类型的infoException。

我们可以指定resttemplate根据Http响应代码解组输出吗？

Answer 1

其余模板在提取数据之前检查错误响应。因此，一种方法是使用ResponseErrorHandler的实现创建Rest模板并覆盖handleError方法以使用转换器提取对象的错误响应，然后将此错误对象包装到异常中。像这样的东西。

    class ErrorObjectResponseHandler implements ResponseErrorHandler {

      private List<HttpMessageConverter<?>> messageConverters;

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

      protected boolean hasError(HttpStatus statusCode) {
        return (statusCode.is4xxClientError() || statusCode.is5xxServerError());
      }

      @Override
      public void handleError(ClientHttpResponse response) throws IOException {
        HttpMessageConverterExtractor<Error> errorMessageExtractor =
          new HttpMessageConverterExtractor(Error.class, messageConverters);
        Error errorObject = errorMessageExtractor.extractData(response);
        throw new SomeException(errorObject);
      }

      public void setMessageConverters(List<HttpMessageConverter<?>> messageConverters) {
        this.messageConverters = messageConverters;
      }

Answer 2

Spring 5专门为此目的提供ExtractingResponseErrorHandler

Answer 3

由于从XML到类的映射，它的XML到POJO无法反序列化。尝试使用此POJO。

public class Info implements Serializable{

    private String TestGuid;

    private String TestUID;

    private String FirstName;

    private String xmlns;

    private String Token;

    private String LastName;

    private String TestUserID;

    //getters and setter
}

在Spring Rest Service调用期间，取消依赖于HTTP代码的解组响应

3 个答案: