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Spring Boot Webservice SOAP错误:内容编码必须设置为gzip

content-type soap web-services spring java

Stimpson Cat · 技术社区 · 6 年前

启动ApplicationContext时出错。若要显示“条件”报告,请在启用“调试”的情况下重新运行应用程序。 2018-12-05 09:57:47 | ERROR | main | work.boot.SpringApplication:858 |应用程序运行失败 java.lang.IllegalStateException:未能执行CommandLineRunner 在org.springframework.boot.SpringApplication.callRunners上(SpringApplication.java:797) 在org.springframework.boot.SpringApplication.run上(SpringApplication.java:324) 在org.springframework.boot.SpringApplication.run上(SpringApplication.java:1260) 在org.springframework.boot.SpringApplication.run上(SpringApplication.java:1248) 在de.varengold.cdl.mvp.MvpApplication.main(MvpApplication.java:21) 内容编码必须设置为gzip

如何将内容编码设置为gzip?

这是我的代码:

@SpringBootApplication

@Slf4j

public class MvpApplication {

public static void main(String[] args) {

SpringApplication.run(MvpApplication.class, args);

}

 @Bean
CommandLineRunner lookup(SOAPConnector soapConnector , Config config) {
return args -> {
  SubmitDATTRA request = new SubmitDATTRA();
  FileDataSource ds = new FileDataSource(config.getReportFilePath() );
  request.setDatei(new DataHandler(ds));
  SubmitDATTRAResponse response =(SubmitDATTRAResponse) soapConnector.callWebService(Config.TEST_URL, request);


    };
  }
}

内容类型是(sysout):application/octet stream。该文件是一个gzip文件。我想我已经接近解决方案了,但是我不知道如何告诉spring应该使用content Encoding=gzip

这是SOAPConnector

import javax.jws.WebService;
import org.springframework.ws.client.core.WebServiceTemplate;
import org.springframework.ws.client.core.support.WebServiceGatewaySupport;

public class SOAPConnector extends WebServiceGatewaySupport {
   public Object callWebService(String url, Object request){
   return getWebServiceTemplate().marshalSendAndReceive(url, request);
   }

}

@Configuration
@Getter
public class Config {

  public static final String PROD_URL = "https://portal.mvp.bafin.de:444/services/ws/a26mifir";
  public static final String TEST_URL = "https://portal.mvp.bafin.de:444/services/ws/t_a26mifir";

  @Value("${report.file.path}")
  private String reportFilePath;

  @Bean
  public Jaxb2Marshaller marshaller() {
    Jaxb2Marshaller marshaller = new Jaxb2Marshaller();
    marshaller.setContextPath("de.acme.ws");
    return marshaller;
  }

  @Bean
  public SOAPConnector soapConnector(Jaxb2Marshaller marshaller) {
    SOAPConnector client = new SOAPConnector();
    client.setDefaultUri(TEST_URL);
    client.setMarshaller(marshaller);
    client.setUnmarshaller(marshaller);
    return client;
  }
}

我读到这篇文章,并从中得到灵感: https://howtodoinjava.com/spring-boot/spring-soap-client-webservicetemplate/