WS客户端:访问本地WSDL的正确途径是什么?

问题是我需要从我提供的文件构建Web服务客户端。 我已经将这个文件存储在本地文件系统中,并且当我将WSDL文件保存在正确的文件系统文件夹中时,一切都很好。 当我将其部署到服务器或从文件系统文件夹中删除WSDL时,代理无法找到WSDL并产生错误。 我搜索了网页,发现了以下文章,但我无法使其工作:
JAX-WS从jar中加载WSDL
http://www.java.net/forum/topic/glassfish/metro-and-jaxb/client-jar-cant-find-local-wsdl-0
http://blog.vinodsingh.com/2008/12/locally-packaged-wsdl.html

我正在使用NetBeans 6.1(这是一个遗留应用程序,我需要使用这个新的Web服务客户端进行更新)。 以下是JAX-WS代理类:

    @WebServiceClient(name = "SOAService", targetNamespace = "http://soaservice.eci.ibm.com/", wsdlLocation = "file:/C:/local/path/to/wsdl/SOAService.wsdl")
public class SOAService
    extends Service
{

    private final static URL SOASERVICE_WSDL_LOCATION;
    private final static Logger logger = Logger.getLogger(com.ibm.eci.soaservice.SOAService.class.getName());

    static {
        URL url = null;
        try {
            URL baseUrl;
            baseUrl = com.ibm.eci.soaservice.SOAService.class.getResource(".");
            url = new URL(baseUrl, "file:/C:/local/path/to/wsdl/SOAService.wsdl");
        } catch (MalformedURLException e) {
            logger.warning("Failed to create URL for the wsdl Location: 'file:/C:/local/path/to/wsdl/SOAService.wsdl', retrying as a local file");
            logger.warning(e.getMessage());
        }
        SOASERVICE_WSDL_LOCATION = url;
    }

    public SOAService(URL wsdlLocation, QName serviceName) {
        super(wsdlLocation, serviceName);
    }

    public SOAService() {
        super(SOASERVICE_WSDL_LOCATION, new QName("http://soaservice.eci.ibm.com/", "SOAService"));
    }

    /**
     * @return
     *     returns SOAServiceSoap
     */
    @WebEndpoint(name = "SOAServiceSOAP")
    public SOAServiceSoap getSOAServiceSOAP() {
        return super.getPort(new QName("http://soaservice.eci.ibm.com/", "SOAServiceSOAP"), SOAServiceSoap.class);
    }

    /**
     * @param features
     *     A list of {@link javax.xml.ws.WebServiceFeature} to configure on the proxy.  Supported features not in the <code>features</code> parameter will have their default values.
     * @return
     *     returns SOAServiceSoap
     */
    @WebEndpoint(name = "SOAServiceSOAP")
    public SOAServiceSoap getSOAServiceSOAP(WebServiceFeature... features) {
        return super.getPort(new QName("http://soaservice.eci.ibm.com/", "SOAServiceSOAP"), SOAServiceSoap.class, features);
    }

}


这是我使用代理的代码:

   WebServiceClient annotation = SOAService.class.getAnnotation(WebServiceClient.class);
   // trying to replicate proxy settings
   URL baseUrl = com.ibm.eci.soaservice.SOAService.class.getResource("");//note : proxy uses "."
   URL url = new URL(baseUrl, "/WEB-INF/wsdl/client/SOAService.wsdl");
   //URL wsdlUrl = this.getClass().getResource("/META-INF/wsdl/SOAService.wsdl"); 
   SOAService serviceObj = new SOAService(url, new QName(annotation.targetNamespace(), annotation.name()));
   proxy = serviceObj.getSOAServiceSOAP();
   /* baseUrl;

   //classescomibmecisoaservice
   //URL url = new URL(baseUrl, "../../../../wsdl/SOAService.wsdl");

   proxy = new SOAService().getSOAServiceSOAP();*/
   //updating service endpoint 
   Map<String, Object> ctxt = ((BindingProvider)proxy ).getRequestContext();
   ctxt.put(JAXWSProperties.HTTP_CLIENT_STREAMING_CHUNK_SIZE, 8192);
   ctxt.put(BindingProvider.ENDPOINT_ADDRESS_PROPERTY, WebServiceUrl);

NetBeans将WSDL的副本放在web-inf / wsdl / client / SOAService中,所以我不想将它添加到META-INF中。 服务类位于WEB-INF / classes / com / ibm / eci / soaservice /中,并且baseurl变量包含文件系统的完整路径(c: path to the project ... soaservice)。 上面的代码引发了这个错误:

javax.xml.ws.WebServiceException:无法在以下位置访问WSDL:file:/WEB-INF/wsdl/client/SOAService.wsdl。 它失败了: WEB-INF wsdl client SOAService.wsdl(找不到路径)

那么,首先,我应该更新代理类的wsdllocation吗? 那么如何告诉WEB-INF / classes / com / ibm / eci / soaservice中的SOAService类来搜索 WEB-INF wsdl client SOAService.wsdl中的WSDL?

编辑 :我找到了这个其他链接 - http://jianmingli.com/wp/?cat=41,它表示将WSDL放入类路径。 我很惭愧地问:我如何将它放入Web应用程序类路径中?


最好的选择是使用jax-ws-catalog.xml

编译本地WSDL文件时,请覆盖WSDL位置并将其设置为类似的内容

http://localhost/wsdl/SOAService.wsdl

不要担心,这只是一个URI而不是一个URL,这意味着您不必在该地址提供WSDL。
您可以通过将wsdllocation选项传递给wsdl来执行此操作以获得java编译器。

这样做会改变你的代理代码

static {
    URL url = null;
    try {
        URL baseUrl;
        baseUrl = com.ibm.eci.soaservice.SOAService.class.getResource(".");
        url = new URL(baseUrl, "file:/C:/local/path/to/wsdl/SOAService.wsdl");
    } catch (MalformedURLException e) {
        logger.warning("Failed to create URL for the wsdl Location: 'file:/C:/local/path/to/wsdl/SOAService.wsdl', retrying as a local file");
        logger.warning(e.getMessage());
    }
    SOASERVICE_WSDL_LOCATION = url;
}

static {
    URL url = null;
    try {
        URL baseUrl;
        baseUrl = com.ibm.eci.soaservice.SOAService.class.getResource(".");
        url = new URL(baseUrl, "http://localhost/wsdl/SOAService.wsdl");
    } catch (MalformedURLException e) {
        logger.warning("Failed to create URL for the wsdl Location: 'http://localhost/wsdl/SOAService.wsdl', retrying as a local file");
        logger.warning(e.getMessage());
    }
    SOASERVICE_WSDL_LOCATION = url;
}

通知文件://在URL构造函数中更改为http://。

现在进入jax-ws-catalog.xml。 没有jax-ws-catalog.xml,jax-ws确实会尝试从该位置加载WSDL

http://localhost/wsdl/SOAService.wsdl
并失败,因为没有这样的WSDL可用。

但是使用jax-ws-catalog.xml,只要它试图访问WSDL @,就可以将jax-ws重定向到本地打包的WSDL,

http://localhost/wsdl/SOAService.wsdl

这里是jax-ws-catalog.xml

<catalog xmlns="urn:oasis:names:tc:entity:xmlns:xml:catalog" prefer="system">
        <system systemId="http://localhost/wsdl/SOAService.wsdl"
                uri="wsdl/SOAService.wsdl"/>
    </catalog>

你正在做的是告诉jax-ws,当它需要加载WSDL时

http://localhost/wsdl/SOAService.wsdl
它应该从本地路径wsdl / SOAService.wsdl加载它。

现在应该在哪里放置wsdl / SOAService.wsdl和jax-ws-catalog.xml? 这是百万美元的问题,不是吗?
它应该位于应用程序jar的META-INF目录中。

所以像这样的东西

ABCD.jar  
|__ META-INF    
    |__ jax-ws-catalog.xml  
    |__ wsdl  
        |__ SOAService.wsdl  

这样,您甚至不必覆盖访问代理的客户端中的URL。 WSDL是从你的JAR中挑选出来的,并且避免在你的代码中使用硬编码的文件系统路径。

有关jax-ws-catalog.xml的更多信息http://jax-ws.java.net/nonav/2.1.2m1/docs/catalog-support.html

希望有所帮助


我们成功采用的另一种方法是使用wsimport(从Ant作为Ant任务)生成WS客户端代理代码,并指定wsdlLocation属性。

<wsimport debug="true" keep="true" verbose="false" target="2.1" sourcedestdir="${generated.client}" wsdl="${src}${wsdl.file}" wsdlLocation="${wsdl.file}">
</wsimport>

由于我们为具有多个WSDL的项目运行此脚本,因此脚本将动态解析$(wsdl.file)值,该值设置为/META-INF/wsdl/YourWebServiceName.wsdl相对于JavaSource位置(或/ src,取决于你如何设置项目)在构建过程中,WSDL和XSD文件被复制到这个位置并打包到JAR文件中(类似于上面Bhasakar描述的解决方案)

MyApp.jar
|__META-INF
   |__wsdl
      |__YourWebServiceName.wsdl
      |__YourWebServiceName_schema1.xsd
      |__YourWebServiceName_schmea2.xsd

注意:确保WSDL文件对所有导入的XSD使用相对引用,而不是http URL:

  <types>
    <xsd:schema>
      <xsd:import namespace="http://valueobject.common.services.xyz.com/" schemaLocation="YourWebService_schema1.xsd"/>
    </xsd:schema>
    <xsd:schema>
      <xsd:import namespace="http://exceptions.util.xyz.com/" schemaLocation="YourWebService_schema2.xsd"/>
    </xsd:schema>
  </types>

生成的代码中,我们发现这个:

/**
 * This class was generated by the JAX-WS RI.
 * JAX-WS RI 2.2-b05-
 * Generated source version: 2.1
 * 
 */
@WebServiceClient(name = "YourService", targetNamespace = "http://test.webservice.services.xyz.com/", wsdlLocation = "/META-INF/wsdl/YourService.wsdl")
public class YourService_Service
    extends Service
{

    private final static URL YOURWEBSERVICE_WSDL_LOCATION;
    private final static WebServiceException YOURWEBSERVICE_EXCEPTION;
    private final static QName YOURWEBSERVICE_QNAME = new QName("http://test.webservice.services.xyz.com/", "YourService");

    static {
        YOURWEBSERVICE_WSDL_LOCATION = com.xyz.services.webservice.test.YourService_Service.class.getResource("/META-INF/wsdl/YourService.wsdl");
        WebServiceException e = null;
        if (YOURWEBSERVICE_WSDL_LOCATION == null) {
            e = new WebServiceException("Cannot find '/META-INF/wsdl/YourService.wsdl' wsdl. Place the resource correctly in the classpath.");
        }
        YOURWEBSERVICE_EXCEPTION = e;
    }

    public YourService_Service() {
        super(__getWsdlLocation(), YOURWEBSERVICE_QNAME);
    }

    public YourService_Service(URL wsdlLocation, QName serviceName) {
        super(wsdlLocation, serviceName);
    }

    /**
     * 
     * @return
     *     returns YourService
     */
    @WebEndpoint(name = "YourServicePort")
    public YourService getYourServicePort() {
        return super.getPort(new QName("http://test.webservice.services.xyz.com/", "YourServicePort"), YourService.class);
    }

    /**
     * 
     * @param features
     *     A list of {@link javax.xml.ws.WebServiceFeature} to configure on the proxy.  Supported features not in the <code>features</code> parameter will have their default values.
     * @return
     *     returns YourService
     */
    @WebEndpoint(name = "YourServicePort")
    public YourService getYourServicePort(WebServiceFeature... features) {
        return super.getPort(new QName("http://test.webservice.services.xyz.com/", "YourServicePort"), YourService.class, features);
    }

    private static URL __getWsdlLocation() {
        if (YOURWEBSERVICE_EXCEPTION!= null) {
            throw YOURWEBSERVICE_EXCEPTION;
        }
        return YOURWEBSERVICE_WSDL_LOCATION;
    }

}

也许这也可能有帮助。 这只是一种不使用“目录”方法的不同方法。


非常感谢Bhaskar Karambelkar的回答,详细解释并解决了我的问题。 但是我也想用三个简单的步骤来回答那些急于解决问题的人的答案

  • 使您的wsdl本地位置参考为wsdlLocation= "http://localhost/wsdl/yourwsdlname.wsdl"
  • 在src下创建一个META-INF文件夹。 将您的wsdl文件放在META-INF下的文件夹中,比如说META-INF / wsdl
  • 在META-INF下创建一个xml文件jax-ws-catalog.xml,如下所示

    <catalog xmlns="urn:oasis:names:tc:entity:xmlns:xml:catalog" prefer="system"> <system systemId="http://localhost/wsdl/yourwsdlname.wsdl" uri="wsdl/yourwsdlname.wsdl" /> </catalog>

  • 现在打包你的罐子。 没有更多的参考本地目录,它都被打包并引用

    链接地址: http://www.djcxy.com/p/66653.html

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