Tomcat7部署中的addServlet错误

Question

我使用以下方法创建了一个 Web 应用程序： - HTML 5 - WebSocket（使用 Tomcat 7） - 春天 - 休眠

问题在于我在 Tomcat 7 中部署战争期间：

org.apache.catalina.startup.ContextConfig parseWebXml
Grave: Erreur d'Úvaluation (parse) dans le fichier web.xml de l'application Ó jn
di:/localhost/mywebapp/WEB-INF/web.xml
org.xml.sax.SAXParseException; systemId: jndi:/localhost/mywebapp/WEB-INF/web.xm
l; lineNumber: 11; columnNumber: 12; Error at (11, 12) : org.apache.catalina.dep
loy.WebXml addServlet

这里是我的 web.xml：

<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/javaee"    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">

<display-name>My WebApp</display-name>

<servlet>
    <servlet-name>Welcome</servlet-name>
    <servlet-class>websocket.UrlWebSocketServlet</servlet-class>
</servlet>

<servlet-mapping>
    <servlet-name>Welcome</servlet-name>
    <url-pattern>/websocket/welcome</url-pattern>
</servlet-mapping>

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

<context-param>
    <param-name>contextConfigLocation</param-name>
    <param-value>/WEB-INF/applicationContext*.xml</param-value>
</context-param>
<listener>
    <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

</web-app>

这里是我的 Servlet（WebSocket 类）：

public class UrlWebSocketServlet extends WebSocketServlet {

private static final Logger log = LoggerFactory.getLogger(UrlWebSocketServlet.class);

@Override
protected boolean verifyOrigin(String origin) {
    log.trace("Origin: {}", origin);
    return true;
}

@Override
protected StreamInbound createWebSocketInbound(String subProtocol, HttpServletRequest request) {
    return new WebSocketConnection();
}

private static class WebSocketConnection extends MessageInbound {

    @Override
    protected void onOpen(WsOutbound outbound) {
        log.info("Conexión abierta");
    }

    @Override
    protected void onClose(int status) {
        log.info("Conexión cerrada");
    }

    @Override
    protected void onBinaryMessage(ByteBuffer byteBuffer) throws IOException {
        log.warn("No se soportan mensajes binarios");
        throw new UnsupportedOperationException("No se soportan mensajes binarios");
    }

    @Override
    protected void onTextMessage(CharBuffer charBuffer) throws IOException {
        final String user = charBuffer.toString();
        log.debug("Mensaje recibido: {}", user);
        getWsOutbound().writeTextMessage(CharBuffer.wrap("Hola " + user + " desde WebSocket"));
    }
}
}

编辑：

这里是我 pom.xml 的 Tomcat jar 部分：

    <dependency>
        <groupId>org.apache.tomcat</groupId>
        <artifactId>tomcat-catalina</artifactId>
        <version>7.0.29</version>
    </dependency>
    <dependency>
        <groupId>org.apache.tomcat</groupId>
        <artifactId>tomcat-coyote</artifactId>
        <version>7.0.29</version>
        <scope>provided</scope>
    </dependency>
    <dependency>
        <groupId>org.apache.tomcat</groupId>
        <artifactId>tomcat-servlet-api</artifactId>
        <version>7.0.29</version>
        <scope>provided</scope>
    </dependency>
            <dependency>
        <groupId>javax.servlet</groupId>
        <artifactId>javax.servlet-api</artifactId>
        <version>3.0.1</version>
        <scope>provided</scope>
    </dependency>

非常感谢:)

克里斯

Answer 1

<dependency>
    <groupId>org.apache.tomcat</groupId>
    <artifactId>tomcat-catalina</artifactId>
    <version>7.0.29</version>
</dependency>

应该说：

<dependency>
    <groupId>org.apache.tomcat</groupId>
    <artifactId>tomcat-catalina</artifactId>
    <version>7.0.29</version>
    <scope>provided</scope>
</dependency>