Question

我使用wsimport gradle任务生成了一个初步的MyService，并提供了wsdl位置路径文件：/ 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;
    }
}

我该如何更改？这是因为文件是在一个环境中生成的，然后工件（war）被移动到另一个服务器。

有什么想法吗？

是的，我明白了。本地一切都很完美。但是这个文件位于war文件中，当Jenkins试图获取这个文件/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）。看起来它无法在war文件中看到文件。有什么想法我怎么处理这个？

Answer 1

使用服务类MyService的构造函数传递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。这是解决方案
@Value("classpath:com//google//resources//wsdl//myservice.interface.v2.wsdl") public void setWsdlLocation(final Resource wsdlLocation) { m_wsdlLocation = wsdlLocation; }

如何更改服务中的wsdl位置文件

2 个答案: