java - Servlet 类 com.foobar.jaxrs.application.MyApplication 不是 javax.servlet.Servlet

Question

我正在尝试使用 Servlet 3.1 和应用程序子类配置 Jersey。阅读文档已经有一段时间了，并试图让它继续下去，但我不确定这里出了什么问题。

web.xml(虽然我不应该需要它，但没有它我就会得到 404...)

<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
id="WebApp_ID" version="3.1">

<display-name>Foobar Models</display-name>

<welcome-file-list>
    <welcome-file>index.htm</welcome-file>
</welcome-file-list>

<session-config>
    <session-timeout>30</session-timeout>
    <cookie-config>
        <name>SESSIONID</name>
    </cookie-config>
</session-config>

<servlet>
    <servlet-name>Foo Bar Application</servlet-name>
    <servlet-class>com.foobar.jaxrs.application.FooBarApplication</servlet-class>
</servlet>

<servlet-mapping>
    <servlet-name>Foo Bar Application</servlet-name>
    <url-pattern>/api/*</url-pattern>
</servlet-mapping>

</web-app>

应用程序子类

package com.foobar.jaxrs.application;

import java.util.HashSet;
import java.util.Set;

import javax.ws.rs.ApplicationPath;
import javax.ws.rs.core.Application;

@ApplicationPath("/api")
public class FooBarApplication extends Application {
public Set<Class<?>> getClasses() {
    Set<Class<?>> s = new HashSet<Class<?>>();
    s.add(com.foobar.api.HealthCheckResource.class);
    return s;
}
}

HealthCheckResource.java

package com.foobar.api;

import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;

@Path("health")
public class HealthCheckResource {

    @GET
    @Produces("text/html")
    public String getHeath() {
        return "Foo Bar Application is healthy!";
    }

}

在jetty中运行(与Tomcat 8相同)

HTTP ERROR 404

Problem accessing /foobar/api/health. Reason:

    Servlet class com.foobar.jaxrs.application.FooBarApplication is not a javax.servlet.Servlet
Caused by:

javax.servlet.UnavailableException: Servlet class com.foobar.jaxrs.application.FooBarApplication is not a javax.servlet.Servlet
    at org.mortbay.jetty.servlet.ServletHolder.checkServletType(ServletHolder.java:362)
    at org.mortbay.jetty.servlet.ServletHolder.doStart(ServletHolder.java:243)
    at org.mortbay.component.AbstractLifeCycle.start(AbstractLifeCycle.java:50)
    at org.mortbay.jetty.servlet.ServletHandler.initialize(ServletHandler.java:685)
....

Answer 1

嗯，消息说明了一切。您正在尝试将应用程序子类部署为 servlet，该子类不是从 Servlet 扩展的。这不可能行得通。

这不是the Jersey documentation告诉做事。它是这样说的:

Hooking up Jersey as a Servlet

<web-app>
    <servlet>
        <servlet-name>MyApplication</servlet-name>
        <servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
        <init-param>
            ...
        </init-param>
    </servlet>
    ...
    <servlet-mapping>
        <servlet-name>MyApplication</servlet-name>
        <url-pattern>/myApp/*</url-pattern>
    </servlet-mapping>
    ...
</web-app>

请注意，servlet 类是org.glassfish.jersey.servlet.ServletContainer。不是您的自定义应用程序子类。

Custom Application subclass

If you extend the Application class to provide the list of relevant root resource classes (getClasses()) and singletons (getSingletons()), i.e. your JAX-RS application model, you then need to register it in your web application web.xml deployment descriptor using a Servlet or Servlet filter initialization parameter with a name of javax.ws.rs.Application [sic] as follows:

Example 4.11. Configuring Jersey container Servlet or Filter to use custom Application subclass

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

请注意，自定义应用程序子类被配置为 servlet 的 init-param。不作为 servlet 类。