如何更改内部供水

Question

WSDL位置的文件我用提供的WSDL位置路径文件中的wsimport任务的gradle产生了初步的MyService：/ d：/someLocationWherePlacedMyWSDl.interface.v2.wsdl

public class MyService 
    extends Service 
{ 

    private final static URL MyService_WSDL_LOCATION; 
    private final static Logger logger = Logger.getLogger(com.google.services.MyService.class.getName()); 

    static { 
     URL url = null; 
     try { 
      URL baseUrl; 
      baseUrl = com.google.services.MyService.class.getResource("."); 
      url = new URL(baseUrl, "file:/D:/someLocationWherePlacedMyWSDl.interface.v2.wsdl"); 
     } catch (MalformedURLException e) { 
      logger.warning("Failed to create URL for the wsdl Location: 'file:/D:/someLocationWherePlacedMyWSDl.interface.v2.wsdl', retrying as a local file"); 
      logger.warning(e.getMessage()); 
     } 
     MyService_WSDL_LOCATION = url; 
    } 
} 

我怎样才能更改？它发生是因为该文件是在一个环境中生成的，然后该工件（战争）被移动到另一个服务器。

有什么想法？

---

是的，我明白了。本地一切都完美。但是这个文件位于war文件内部，当Jenkins试图获得这个文件/var/distributives/myservice/tomcat-base/wsdl/someLocationWherePlacedMyWSDl.interface.v2.wsdl时，我得到异常（没有这样的文件或目录）。它看起来像无法看到战争文件内的文件。任何想法如何处理？

Answer 1

使用您的服务类的构造函数MyService来传递wsdlLocation。

String WSDL_LOCATION = "http://server:port/localtionWSDL.interface.v2.wsdl"; 

try { 
    final URL url = new URL(WSDL_LOCATION); 
    final QName serviceName = new QName("http://mynamespace/", "MyService"); 
    final MyService service = new MyService(url, serviceName); 
    port = service.getMyServicePort(); 

    // Call some operation of WebService 

} catch (final Exception e) { 
    // Handle the exception 
} 

Answer 2

我用相对路径解决了这个问题。这里是解决方案
@Value("classpath:com//google//resources//wsdl//myservice.interface.v2.wsdl") public void setWsdlLocation(final Resource wsdlLocation) { m_wsdlLocation = wsdlLocation; }