將多個servlet添加到一個web.xml中

Question

我想將澤西島項目B（已經運行良好）合併/添加到新澤西項目A中，該項目將充當過濾器/安全層。因此，作爲一個基本步驟，我在項目A的構建路徑中向項目B添加了依賴項，並將其添加到構建路徑中的部署程序集。我從this的帖子瞭解到，我可以通過使servlet在相同的web.xml中並使用<servlet-mapping>以不同方式映射它們來實現。我沒有任何運氣，當我試圖訪問B項目的資源

web.xml

<servlet> 
    <servlet-name>Jersey Web Application</servlet-name> 
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class> 
    <init-param> 
     <param-name>jersey.config.server.provider.packages</param-name> 
     <param-value>org.abc.def.ba, org.pqr.xyz</param-value> 
    </init-param> 
    <init-param> 
     <param-name>javax.ws.rs.Application</param-name> 
     <param-value>org.abc.def.ba.CustomApplication</param-value> 
    </init-param> 
    <load-on-startup>1</load-on-startup> 
</servlet> 

<servlet-mapping> 
    <servlet-name>Jersey Web Application</servlet-name> 
    <url-pattern>/webapi/*</url-pattern> 
</servlet-mapping> 

<!-- The Servlet of Project B --> 
<servlet> 
    <servlet-name>Jersey Web Application 2</servlet-name> 
    <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-  class> 

<init-param> 
     <param-name>javax.ws.rs.Application</param-name> 
     <param-value> org.pqr.xyz.MyApplication</param-value> 
    </init-param> 

    <!-- Register resources and providers under my.package. --> 
    <init-param> 
     <param-name>jersey.config.server.provider.packages</param-name> 
     <param-value> org.pqr.xyz</param-value> 
    </init-param> 

    <!-- Register my custom provider (not needed if it's in my.package) AND LoggingFilter. --> 
    <init-param> 
     <param-name>jersey.config.server.provider.classnames</param-name> 
     <param-value> org.pqr.xyz.mapper.ObjectMapperProvider</param-value> 
    </init-param> 

    <init-param> 
     <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name> 
     <param-value>true</param-value> 
    </init-param> 
    <init-param> 
     <param-name>com.sun.jersey.config.property.packages</param-name> 
     <param-value> org.pqr.xyzr</param-value> 
    </init-param> 
    <init-param> 
     <param-name>com.sun.jersey.spi.container.ResourceFilters</param-name> 
     <param-value>com.porterhead.rest.filter.ResourceFilterFactory</param-value> 
    </init-param> 
    <init-param> 
     <param-name>readOnly</param-name> 
     <param-value>false</param-value> 
    </init-param> 
    <load-on-startup>2</load-on-startup> 
</servlet> 
<servlet-mapping> 
    <servlet-name>Jersey Web Application 2</servlet-name> 
    <url-pattern>/rest/*</url-pattern> 
</servlet-mapping> 

所以，當我嘗試訪問http://localhost:8080/ba/webapi/myresource，它工作得很好。但是，當我嘗試http://localhost:8080/ba/rest/newresource時，出現404錯誤。我知道我錯過了一些東西。我感謝你的指導。

Answer 1

如果您使用的是servlet 3.0，那麼您可以使用@ApplicationPath註釋爲不同的資源使用不同的應用程序路徑。如果你不想使用註釋，就可以實現使用的web.xml下同

<web-app> 
<servlet> 
    <servlet-name>org.foo.rest.MyApplication</servlet-name> 
</servlet> 
... 
<servlet-mapping> 
    <servlet-name>org.foo.rest.MyApplication</servlet-name> 
    <url-pattern>/resources</url-pattern> 
</servlet-mapping> 
... 

詳見新澤西documentation的部署選項。