ServerEndpoint 和 web.xml

Question

我有一些 Soap、REST servlet，现在有一个 WebSocket：

@ServerEndpoint("/game")
public class WebSocketgame{
...
}

我有下一个麻烦：如果 web.xml 存在，WebSocket 不可见。在 web.xml 中描述 jdbc 资源、servlets 和其他... 当我删除 web.xml - websocket 成功可见。我该如何解决这个问题？

更新 web.xml：

<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">
    <listener>
        <listener-class>com.sun.xml.ws.transport.http.servlet.WSServletContextListener</listener-class>
    </listener>
    <servlet>
        <servlet-name>ConfigServlet</servlet-name>
        <servlet-class>com.example.ConfigServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet>
        <servlet-name>MainService</servlet-name>
        <servlet-class>com.sun.xml.ws.transport.http.servlet.WSServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>ConfigServlet</servlet-name>
        <url-pattern>/ConfigServlet</url-pattern>
    </servlet-mapping>

    <servlet>
        <servlet-name>JsonServlet</servlet-name>
        <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
        <init-param>
            <param-name>com.sun.jersey.config.property.packages</param-name>
            <param-value>com.example.json</param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>JsonServlet</servlet-name>
        <url-pattern>/json/*</url-pattern>
    </servlet-mapping>
    <context-param>
        <param-name>propfile</param-name>
        <param-value>/WEB-INF/server_config.txt</param-value>
    </context-param>
    <servlet-mapping>
        <servlet-name>MainService</servlet-name>
        <url-pattern>/MainService</url-pattern>
    </servlet-mapping>
    <session-config>
        <session-timeout>
            30
        </session-timeout>
    </session-config>
    <resource-ref>
        <description>postgreSQL Datasource example</description>
        <res-ref-name>jdbc/postgres</res-ref-name>
        <res-type>javax.sql.DataSource</res-type>
        <res-auth>Container</res-auth>
    </resource-ref>
</web-app>

Answer 1

尝试添加

<beans xmlns="http://xmlns.jcp.org/xml/ns/javaee"
   xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
   xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/beans_1_1.xsd"
   bean-discovery-mode="annotated">

在 beans.xml 文件中

Answer 2

我相信你需要为那些有这种情况的人添加 servlet 配置

<servlet>
<servlet-name>gameServer</servlet-name>
<servlet-class>WebSocketgame.SocketEndPoint</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>gameServer</servlet-name>
<url-pattern>/game</url-pattern>
</servlet-mapping>

Answer 3

我认为您应该配置您的主 servlet (MainService) 以提供所有 url /* 或 /MainService/*，而不是使用 websocket 作为 /MainService/game。

你的班级

@ServerEndpoint("/MainService/game")
public class WebSocketgame{
...
}

还有 web.xml http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"> com.sun.xml.ws.transport.http.servlet.WSServletContextListener ConfigServlet com.example.ConfigServlet 1 主要服务 com.sun.xml.ws.transport.http.servlet.WSServlet 1

<servlet-mapping>
    <servlet-name>ConfigServlet</servlet-name>
    <url-pattern>/ConfigServlet</url-pattern>
</servlet-mapping>

<servlet>
    <servlet-name>JsonServlet</servlet-name>
    <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
    <init-param>
        <param-name>com.sun.jersey.config.property.packages</param-name>
        <param-value>com.example.json</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>JsonServlet</servlet-name>
    <url-pattern>/json/*</url-pattern>
</servlet-mapping>
<context-param>
    <param-name>propfile</param-name>
    <param-value>/WEB-INF/server_config.txt</param-value>
</context-param>
<servlet-mapping>
    <servlet-name>MainService</servlet-name>
    <url-pattern>/MainService/*</url-pattern>
</servlet-mapping>
<session-config>
    <session-timeout>
        30
    </session-timeout>
</session-config>
<resource-ref>
    <description>postgreSQL Datasource example</description>
    <res-ref-name>jdbc/postgres</res-ref-name>
    <res-type>javax.sql.DataSource</res-type>
    <res-auth>Container</res-auth>
</resource-ref>

并使您的 js（或其他 websocket 客户端）连接到 /MainService/game 而不是 /game

Answer 4

我遇到了同样的问题。我的解决方案是将端点类扩展到 HttpServlet。

@ServerEndpoint("/game")
public class WebSocketgame extends HttpServlet {
...
}

对于 web.xml 的定义

<servlet>
        <display-name>Webgame</display-name>        
        <servlet-name>Webgame</servlet-name>        
        <servlet-class>com.example.WebSocketgame</servlet-class>        
        <load-on-startup>1</load-on-startup>
    </servlet>