Question

尝试访问/Example/Site时，将显示404找不到的页面，但jsp位于webapp/WEB-INF/jsp/。为什么这不起作用？

<web-app id="WebApp_ID" version="2.4"
         xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
         http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
    <display-name>jmattheis rest app</display-name>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>classpath:applicationContext.xml</param-value>
    </context-param>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>


    <servlet>
        <servlet-name>jersey-servlet</servlet-name>
        <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
        <init-param>
            <param-name>jersey.config.server.provider.packages</param-name>
            <param-value>com.stackoverflow.jmattheis.rest</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.servlet.filter.forwardOn404</param-name>
            <param-value>true</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.server.provider.classnames</param-name>
            <param-value>org.glassfish.jersey.server.mvc.jsp.JspMvcFeature</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.server.mvc.templateBasePath.jsp</param-name>
            <param-value>/WEB-INF/jsp</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.server.tracing</param-name>
            <param-value>ALL</param-value>
        </init-param>
    </servlet>

    <servlet-mapping>
        <servlet-name>jersey-servlet</servlet-name>
        <url-pattern>/*</url-pattern>
    </servlet-mapping>

</web-app>

资源：

@Component
@Path("Example")
public class Example {
    @GET
    @Path("/Site")
    @Produces(MediaType.TEXT_HTML)
    public Response getSite() {
        return Response.ok(new Viewable("/test")).build();
    }
}

Answer 1

Have a look at the Jersey2 MVC documentation

想要使用JSP模板支持的Jersey Web应用程序应该在应用程序的web.xml中注册为Servlet过滤器而不是Servlet。对于需要使用Jersey MVC模板支持的Web应用程序，目前不支持Servlet 3.0中引入的无web.xml部署样式。

因此，您需要将servlet更改为过滤器以使其正常工作。

可能是在早期版本的球衣2中不需要这个，但是我找不到那些旧的文档，所以我们永远不会知道。

像这样：

<web-app id="WebApp_ID" version="2.4"
         xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
         xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
         http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
    <display-name>jmattheis rest app</display-name>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>classpath:applicationContext.xml</param-value>
    </context-param>

    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>


    <filter>
        <filter-name>jersey-filter</filter-name>
        <filter-class>org.glassfish.jersey.servlet.ServletContainer</filter-class>
        <init-param>
            <param-name>jersey.config.server.provider.packages</param-name>
            <param-value>com.stackoverflow.jmattheis.rest</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.servlet.filter.forwardOn404</param-name>
            <param-value>true</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.server.provider.classnames</param-name>
            <param-value>org.glassfish.jersey.server.mvc.jsp.JspMvcFeature</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.server.mvc.templateBasePath.jsp</param-name>
            <param-value>/WEB-INF/jsp</param-value>
        </init-param>
        <init-param>
            <param-name>jersey.config.server.tracing</param-name>
            <param-value>ALL</param-value>
        </init-param>
    </filter>

    <filter-mapping>
        <filter-name>jersey-filter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>

</web-app>

可视路线返回404

1 个答案: