Question

这是我的web.xml

<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_2_5.xsd"
    version="2.5">

    <display-name>XYZ_POC</display-name>

    <servlet>
        <servlet-name>dispatcher</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>

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

    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>/WEB-INF/dispatcher-servlet.xml</param-value>
    </context-param>

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

和我的dispatcher-servlet.xml

<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="
        http://www.springframework.org/schema/beans     
        http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context 
        http://www.springframework.org/schema/context/spring-context.xsd">

    <context:component-scan base-package="com.abc.pqr" />

    <bean
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix">
            <value>/WEB-INF/safePages/</value>
        </property>
        <property name="suffix">
            <value>.jsp</value>
        </property>
    </bean>

</beans>

和我的控制器类

@Controller
public class MainController {

    private static final String VIEW_INDEX = "index";

    @RequestMapping(value = "/", method = RequestMethod.GET)
    public String welcome(ModelMap model) {
        User requestForm = new User();
        model.addAttribute("requestForm",requestForm);
        return VIEW_INDEX;
    }
}

这很好用。

我的jsp文件中有一些CSS和JS代码，我想从JSP中分离出来。

我按照tutorial逐字逐句地执行此操作。

所以我的新调度程序servlet xml变成了：

<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:mvc="http://www.springframework.org/schema/mvc"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="
        http://www.springframework.org/schema/beans     
        http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/mvc 
        http://www.springframework.org/schema/mvc/spring-mvc.xsd
        http://www.springframework.org/schema/context 
        http://www.springframework.org/schema/context/spring-context.xsd">

    <context:component-scan base-package="com.abc.pqr" />

    <bean
        class="org.springframework.web.servlet.view.InternalResourceViewResolver">
        <property name="prefix">
            <value>/WEB-INF/safePages/</value>
        </property>
        <property name="suffix">
            <value>.jsp</value>
        </property>
    </bean>

    <mvc:resources mapping="/resources/theme/**" location="/resources/theme/"/>
    <mvc:resources mapping="/resources/jscript/**" location="/resources/jscript/"/>

</beans>

但是当我部署它时，我的应用程序停止工作。在从localhost访问上下文根时，我开始得到＆＃34;没有找到HTTP请求的映射...＆＃34;错误。

我错过了什么？

Answer 1

我遇到了同样的问题。并在spring spring中添加以下标签为我工作

“MVC：默认servlet的处理程序/＆gt;” 中

使用MVC标记添加css，js时，没有找到带有URI Spring MVC的HTTP请求的映射

1 个答案: