JAX-WS 客户机:访问本地 WSDL 的正确路径是什么?

2022-08-31 11:09:40

问题是我需要从我提供的文件中构建一个Web服务客户端。我已将此文件存储在本地文件系统上,虽然我将 WSDL 文件保存在正确的文件系统文件夹中,但一切都很好。当我将其部署到服务器或从文件系统文件夹中删除 WSDL 时,代理找不到 WSDL 并引发错误。我已经搜索了网络,我找到了以下帖子,但我无法使它工作:
JAX-WS从jar
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

我使用的是 NetBeans 6.1(这是一个旧版应用程序,我必须使用这个新的 Web 服务客户端进行更新)。下面是 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;

   //classes\com\ibm\eci\soaservice
   //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 在 web-inf/wsdl/client/SOAService 中放置了 WSDL 的副本,因此我也不想将其添加到 META-INF 中。服务类位于 WEB-INF/classes/com/ibm/eci/soaservice/ 中,baseurl 变量包含它的文件系统完整路径 (c:\path\to\the\project...\soaservice )。上面的代码引发错误:

javax.xml.ws.WebServiceException:未能访问 WSDL,网址为:file:/WEB-INF/wsdl/client/SOAService.wsdl。失败的原因是:\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,我如何告诉 WEB-INF/classes/com/ibm/eci/soaservice 中的 SOAService 类在 \WEB-INF\wsdl\client\SOAService.wsdl 中搜索 WSDL?

编辑:我发现了另一个链接 - http://jianmingli.com/wp/?cat=41,它说要将WSDL放入类路径中。我很惭愧地问:如何将其放入Web应用程序类路径中?


答案 1

最好的选择是使用 jax-ws-catalog.xml

编译本地 WSDL 文件 时,将覆盖 WSDL 位置并将其设置为类似

http://localhost/wsdl/SOAService.wsdl

不用担心这只是一个URI,而不是一个URL,这意味着您不必在该地址上提供WSDL。
您可以通过将 wsdllocation 选项传递给 wsdl to java 编译器来执行此操作。

这样做会将您的代理代码从

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;
}

请注意,file:// URL 构造函数中更改为 http://。

现在出现在jax-ws-catalog.xml。如果没有 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每当 jax-ws 尝试访问 WSDL @

http://localhost/wsdl/SOAService.wsdl
.

这是 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,当它需要从中加载WSDL时

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

现在你应该把 wsdl/SOAService.wsdl 和 jax-ws-catalog 放在哪里.xml?这就是百万美元的问题,不是吗?
它应该在应用程序 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 JAR 中选取的,您不必在代码中使用硬编码的文件系统路径。

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

希望有所帮助


答案 2

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 的项目运行此函数,因此脚本会动态解析 $(wsdl.file} 值,该值设置为相对于 JavaSource 位置(或 /src,具体取决于您设置项目的方式)的 /META-INF/wsdl/YourWebServiceName.wsdl。在构建过程中,WSDL 和 XSDs 文件将复制到此位置并打包到 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:请确保 WSDL 文件使用与任何导入的 XSD 相对折射,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:

/**
 * 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;
    }

}

也许这也会有所帮助。这只是一种不同的方法,不使用“目录”方法。


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