Spring在tomcat上部署war文件而不显示页面

时间:2014-05-28 09:50:37

标签: spring tomcat

您好我为spring创建了一个maven项目。在我创建项目后,最终产品是.war文件。 然后我使用tomcat manager上传它。但是我无法访问或显示页面...... !! 它给了我以下错误 enter image description here

的web.xml

<!-- Reads request input using UTF-8 encoding -->
    <filter>
        <filter-name>characterEncodingFilter</filter-name>
        <filter-class>org.springframework.web.filter.CharacterEncodingFilter</filter-class>
        <init-param>
            <param-name>encoding</param-name>
            <param-value>UTF-8</param-value>
        </init-param>
        <init-param>
            <param-name>forceEncoding</param-name>
            <param-value>true</param-value>
        </init-param>
    </filter>

    <filter-mapping>
        <filter-name>characterEncodingFilter</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/spring-context-data.xml</param-value>
    </context-param>

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

    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>
            /WEB-INF/spring-context-web.xml,
             /WEB-INF/spring-context-data.xml,
             /WEB-INF/spring-security.xml
            </param-value>

        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
             /WEB-INF/spring-context-web.xml,
             /WEB-INF/spring-context-data.xml,
             /WEB-INF/spring-security.xml
        </param-value>
    </context-param>

    <servlet-mapping>
        <servlet-name>dispatcher</servlet-name>
        <url-pattern>/*</url-pattern>
    </servlet-mapping>
    <!-- <servlet>
            <servlet-name>RealEstate</servlet-name>
            <servlet-class>com.Servlet1</servlet-class>
    </servlet>
    <servlet-mapping>
        <servlet-name>RealEstate</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping> -->
<!-- Handles all requests into the application -->
    <servlet>
        <servlet-name>Spring MVC Dispatcher Servlet</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <init-param>
            <param-name>contextConfigLocation</param-name>
            <param-value>
                /WEB-INF/spring-context-web.xml
            </param-value>
        </init-param>
        <load-on-startup>1</load-on-startup>
    </servlet>

    <servlet-mapping>
        <servlet-name>Spring MVC Dispatcher Servlet</servlet-name>
        <url-pattern>/</url-pattern>
    </servlet-mapping>
    <filter>

        <filter-name>methodFilter</filter-name>
        <filter-class>org.springframework.web.filter.HiddenHttpMethodFilter</filter-class>
    </filter>

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

    <!-- Spring Security -->
    <filter>
        <filter-name>springSecurityFilterChain</filter-name>
        <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
    </filter>

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

    <welcome-file-list>

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

    </welcome-file-list>

的pom.xml

<plugin>
                <artifactId>maven-compiler-plugin</artifactId>
                <version>2.3.2</version>
                <configuration>
                    <source>1.7</source>
                    <target>1.7</target>
                </configuration>
            </plugin>


            <plugin>
                <groupId>org.codehaus.mojo</groupId>
                <artifactId>tomcat-maven-plugin</artifactId>
                <configuration>
                    <url>http://localhost:8080/manager/html</url>
                    <server>tomcat7</server>
                    <path>/RealEstate</path>
                </configuration>
            </plugin>

控制器

@RequestMapping("/home")
    public String getHomePage() {

        return "/index";
    }

2 个答案:

答案 0 :(得分:0)

您的网址格式无法满足您的家庭要求;从下面移除星星:

 <filter-mapping>
    <filter-name>characterEncodingFilter</filter-name>
    <url-pattern>/</url-pattern>
 </filter-mapping>

使用此网址模式适用于每个网址。也适用于其他网址模式。

答案 1 :(得分:0)

您应该发布dispatch servlet xml配置文件(spring-context-web.xml)。你应该像这样配置视图解析器:

<bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="viewClass" value="org.springframework.web.servlet.view.JstlView"/>
        <property name="prefix" value="/WEB-INF/views/" />
        <property name="suffix" value=".jsp" />
        <property name="order" value="2" /> 
</bean>