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 Controller 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,我想将它们与 JSP 分开。
我关注了这个tutorial逐字地去做。
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。从本地主机访问上下文根时,我开始收到“未找到 HTTP 请求的映射...”错误。
我错过了什么?
最佳答案
我遇到了同样的问题。在 spring conf 中添加以下标签对我有用
"mvc:default-servlet-handler/>"
关于javascript - 使用 MVC 标记添加 css、js 时,找不到使用 URI Spring MVC 的 HTTP 请求的映射,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36978437/