使用 SOAP WS 未授权返回错误 401答案

【问题标题】：Consuming SOAP WS returns Error 401 unauthorized使用 SOAP WS 未授权返回错误 401
【发布时间】：2016-04-08 09:30:37
【问题描述】：

我正在尝试使用 spring-ws 使用 SOAP Web 服务。我能够从 SOAP UI 成功使用此 Web 服务。但是我收到一个未经授权的异常：

org.springframework.ws.client.WebServiceTransportException: Unauthorized [401]
    at org.springframework.ws.client.core.WebServiceTemplate.handleError(WebServiceTemplate.java:699)

如何成功验证？

我的代码如下：

WebServiceTemplate 配置：

@Bean
    public WebServiceTemplate webServiceTemplate(Jaxb2Marshaller marshaller){
        WebServiceTemplate webServiceTemplate = new WebServiceTemplate();
        webServiceTemplate.setDefaultUri("http://ultron.illovo.net:9704/AdminService");
        webServiceTemplate.setMarshaller(marshaller);
        webServiceTemplate.setUnmarshaller(marshaller);

        Credentials credentials = new UsernamePasswordCredentials("biqa", "welcome1");

        HttpComponentsMessageSender messageSender = new HttpComponentsMessageSender();
        messageSender.setCredentials(credentials);
        messageSender.setReadTimeout(5000);
        messageSender.setConnectionTimeout(5000);
        webServiceTemplate.setMessageSender(messageSender);

        return webServiceTemplate;
    }

客户：

@Autowired

WebServiceTemplate webServiceTemplate;

public CallProcedureWithResultsResponse callProcedureWithResults(String procedureName){
    CallProcedureWithResults request = new CallProcedureWithResults();
    request.setProcedureName(procedureName);
    log.info("Calling procedure " + procedureName);

    CallProcedureWithResultsResponse response = (CallProcedureWithResultsResponse) webServiceTemplate.marshalSendAndReceive("http://ultron.illovo.net:9704/AdminService/AdminService", request);

    return response;
}

JUnit 测试

@Test
    public void testWebService(){
        AnnotationConfigApplicationContext context = new AnnotationConfigApplicationContext(ObieeConfiguration.class);
        ObieeClient client = context.getBean(ObieeClient.class);
        CallProcedureWithResultsResponse response = client.callProcedureWithResults("GetOBISVersion()");
        client.printResponse(response);
        context.close();
    }

【问题讨论】：

你解决过这个问题吗？如果是这样，您的解决方案是什么？

标签： java web-services soap spring-ws

【解决方案1】：

我通过执行以下操作设法解决了这个问题：

如下创建HttpUrlConnectionMessageSender的子类：

public class WebServiceMessageSenderWithAuth extends HttpUrlConnectionMessageSender {

    String user;
    String password;

    public WebServiceMessageSenderWithAuth(String user, String password) {
        this.user = user;
        this.password = password;
    }

    @Override
     protected void prepareConnection(HttpURLConnection connection) throws IOException { 
        BASE64Encoder enc = new BASE64Encoder(); 
        String userpassword = user+":"+password;
        String encodedAuthorization = enc.encode( userpassword.getBytes() ); 
        connection.setRequestProperty("Authorization", "Basic " + encodedAuthorization); 
        super.prepareConnection(connection); 
    } 

}

我的客户定义如下：

public class ObieeClient extends WebServiceGatewaySupport {

    private static final Logger log = LoggerFactory.getLogger(ObieeClient.class);

    public CallProcedureWithResultsResponse callProcedureWithResults(String procedureName, String webServiceURL){
        CallProcedureWithResults request = new CallProcedureWithResults();
        request.setProcedureName(procedureName);
        log.info("Calling procedure " + procedureName);

        CallProcedureWithResultsResponse response = (CallProcedureWithResultsResponse) ((JAXBElement) getWebServiceTemplate().marshalSendAndReceive(webServiceURL, request)).getValue();

        return response;
    }
}

最后，客户端通过设置消息发送者配置为使用凭据：

     Jaxb2Marshaller marshaller = new Jaxb2Marshaller();            marshaller.setContextPath("za.co.adaptit.smartdeployment.webservices.wsdl");

                //set up web service client
                ObieeClient client = new ObieeClient();
                client.setDefaultUri(host);
                client.setMarshaller(marshaller);
                client.setUnmarshaller(marshaller);
                client.setMessageSender(new WebServiceMessageSenderWithAuth("user", "password"));

response = client.callProcedureWithResults("NQSModifyMetadata('" + obieeObjectsXML +"')", server.getHost());

希望这会有所帮助。

【讨论】：

【解决方案2】：

我遇到了同样的问题，并通过定义带有凭据的 MessageSender 然后将其设置给客户端来设法解决。我不必创建自定义 MessageSender。

@Bean
public WebServiceMessageSender messageSender(Credentials credentials) throws Exception{
    HttpComponentsMessageSender messageSender = new HttpComponentsMessageSender();
    messageSender.setCredentials(credentials);
    return messageSender;
}

@Bean
public Credentials credentials(){
    UsernamePasswordCredentials credentials = new UsernamePasswordCredentials(USERNAME, PASSWORD);
    return credentials;

}

【讨论】：

【解决方案3】：

使用 HttpsUrlConnectionMessageSender 进行安全 Https 连接。这是 spring-ws-support 依赖项的一部分。此外，您可以使用 HttpsUrlConnectonMessageSender 绕过各种证书问题，例如证书公用名不匹配。

不要使用BASE64Encoder，它是sun.misc jar 的类，不太安全。使用 Java 1.8 中存在的 Java util Base64 和 Java 版本低于 Java 1.8 的 Apache 的 Base64

public class WebServiceMessageSenderWithAuth extends HttpsUrlConnectionMessageSender {

String user;
String password;

public WebServiceMessageSenderWithAuth(String user, String password) {
    this.user = user;
    this.password = password;
   }

@Override
 protected void prepareConnection(HttpURLConnection connection) throws IOException { 

    String userpassword = user+":"+password;
    String encodedAuthorization = Base64.encode( userpassword.getBytes() ); 
    connection.setRequestProperty("Authorization", "Basic " + encodedAuthorization); 
    //set various properties by using reference of connection according to your requirement
   } 
}

【讨论】：