3

We are struggling to configure our web app to be able to connect with web services via Spring WS. We have tried to use the example from the documentation of client-side Spring-WS, but we end up with a WebServiceTransportException. The XML config looks like this:

<bean id="webServiceTemplate" class="org.springframework.ws.client.core.WebServiceTemplate">
    <constructor-arg ref="messageFactory"/>
    <property name="messageSender">
        <bean class="org.springframework.ws.transport.http.CommonsHttpMessageSender">
            <property name="credentials">
                <bean class="org.apache.commons.httpclient.UsernamePasswordCredentials">
                    <constructor-arg value="john"/>
                    <constructor-arg value="secret"/>
                </bean>
            </property>
        </bean>
    </property>
</bean>

We have been able to configure the application programmatically, but this configuration was not possible to "transfer" to a Spring XML config because some setters did not use the format Spring expects. (HttpState.setCredentials(...) takes two parameters). The config was lifted from some other Spring-WS client code in the company.

This is the configuration that works:

 public List<String> getAll() {
    List<String> carTypes = new ArrayList<String>();

    try {
        Source source = new ResourceSource(request);
        JDOMResult result = new JDOMResult();


        SaajSoapMessageFactory soapMessageFactory = new SaajSoapMessageFactory(MessageFactory.newInstance());

        WebServiceTemplate template = new WebServiceTemplate(soapMessageFactory);

        HttpClientParams clientParams = new HttpClientParams();
        clientParams.setSoTimeout(60000);
        clientParams.setConnectionManagerTimeout(60000);
        clientParams.setAuthenticationPreemptive(true);

        HttpClient client = new HttpClient(clientParams);
        client.getState().setCredentials(AuthScope.ANY,
                new UsernamePasswordCredentials("username", "password"));

        CommonsHttpMessageSender messageSender = new CommonsHttpMessageSender(client);

        template.setMessageSender(messageSender);
        template.sendSourceAndReceiveToResult(SERVICE_URI,
                source, result);

        // Handle the XML

    } catch (IOException e) {
        throw new RuntimeException(e);
    } catch (SOAPException e) {
        throw new RuntimeException(e);
    }

    return carTypes;
}

Does anyone know how to solve my problem? Every tutorial I have see out there lists the first configuration. It seems that when I set the credentials on the messageSender object, they are just ignored...

4

4 回答 4

2

使用构造函数覆盖 HttpClient，该构造函数接受参数并使用构造函数参数通过 Spring 连接

public MyHttpClient(HttpClientParams params, UsernamePasswordCredentials usernamePasswordCredentials) {
        super(params);        
        getState().setCredentials(AuthScope.ANY, usernamePasswordCredentials);
    }
于 2009-12-16T16:04:01.147 回答
1

你如何区分这些：

<constructor-arg value="john"/>
<constructor-arg value="secret"/>

尝试用这个替换它：

<property name="userName" value="john" />
<property name="password" value="secret" />

希望能帮助到你。

于 2008-09-29T19:10:13.887 回答
0

如果您在示例中使用 defaultHttpClient，请在 HTTPMessageSender 上使用 afterPropertiesSet 方法，这应该可以通过正确应用凭据来解决您的问题

于 2013-05-01T19:41:15.080 回答
0

起初我们在我们的项目中设置凭据是这样的：

<bean id="authenticationEnabledCommonsHttpMessageSender" parent="commonsHttpMessageSender"
    p:credentials-ref="clientCredentials" lazy-init="true" />
<bean id="clientCredentials"
    class="org.apache.commons.httpclient.UsernamePasswordCredentials"
    c:userName="${clientCredentials.userName}"
    c:password="${clientCredentials.password}"
    lazy-init="true" />

这是我们启用凭据的选项。当我们像这样设置凭据时出现问题。如果我们发送消息的服务器（具有 Axis impl）没有获得用户名密码凭据，我们会得到“未经授权”异常。因为，当我们跟踪 vie TCPMon 时，我们意识到“用户名：密码：”字符串被发送，你可以看到用户名和密码没有值。

之后，我们像这样设置凭据：

public Message sendRequest(OutgoingRequest message, MessageHeaders headers,
                        EndpointInfoProvider endpointInfoProvider,
                        WebServiceMessageCallback requestCallback){
    Assert.notNull(endpointInfoProvider, "Destination provider is required!");
    final Credentials credentials = endpointInfoProvider.getCredentials();
    URI destinationUri = endpointInfoProvider.getDestination();
    for (WebServiceMessageSender messageSender : webServiceTemplate.getMessageSenders()) {
        if (messageSender instanceof CommonsHttpMessageSender) {
            HttpClient httpClient = ((CommonsHttpMessageSender) messageSender).getHttpClient();
            httpClient.getState().setCredentials(
                    new AuthScope(destinationUri.getHost(),
                            destinationUri.getPort(), AuthScope.ANY_REALM,
                            AuthScope.ANY_SCHEME), credentials
            );
            httpClient.getParams().setAuthenticationPreemptive(true);
            ((CommonsHttpMessageSender) messageSender)
                    .setConnectionTimeout(endpointInfoProvider
                            .getTimeOutDuration());
        }
    }

getCredentials 方法是：

@Override
public Credentials getCredentials(){
    if (credentials != null) {
        return credentials;
    }
    String username = parameterService.usernameFor(getServiceName());
    String password = parameterService.passwordFor(getServiceName());
    if (username == null && password == null) {
        return null;
    }
    credentials = new UsernamePasswordCredentials(username, password);
    return credentials;
}
于 2014-09-04T08:55:24.047 回答