我已将我的资源分配到界面 - impl因为它更有意义但是如果我让它扫描我的资源,这似乎不受最新版本的球衣的支持。
如何在web.xml中手动定义资源? 如果我在web.xml中手动定义资源impl,这会起作用吗?
感谢 亚历
答案 0 :(得分:1)
[1]。创建一个扩展 javax.ws.rs.core.Application 的java类并注册您的资源:
public class MyRESTApp
extends Application {
@Override
public Set<Class<?>> getClasses() {
Set<Class<?>> s = new HashSet<Class<?>>();
s.add(MyResource.class);
....
return s;
}
}
[2]。在web.xml注册应用程序 如果您正在使用 spring ,请访问web.xml:
<servlet>
<servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.xxx.MyRESTApp</param-value>
</init-param>
...
</servlet>
如果你正在使用 guice ,请访问web.xml:
<filter>
<filter-name>guiceFilter</filter-name>
<filter-class>com.google.inject.servlet.GuiceFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>guiceFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<listener>
<listener-class>com.xx.MyGuiceServletContextListener</listener-class>
</listener>
然后创建扩展 com.google.inject.servlet.GuiceServletContextListener
的java类MyGuiceServletContextListenerpublic class MyGuiceServletContextListener
extends GuiceServletContextListener {
@Override
protected Injector getInjector() {
Guice.createInjector(new JerseyServletModule() {
@Override
protected void configureServlets() {
// Route all requests through GuiceContainer
// IMPORTANT
// If this property is not defined guice tries to find the @Path annotated types defied at the injector
Map<String,String> params = new HashMap<String, String>();
params.put("javax.ws.rs.Application",
MyRESTApp.class.getName());
serve("/*").with(GuiceContainer.class,
params);
});
}
}