我想实现这种模板功能:
@Controller
返回字符串时,它将定义包含在模板像这样:
@Controller
@RequestMapping(value = "/", method = RequestMethod.GET)
public String home(Locale locale, Model model) {
return "home_view";
}
视图/ home_view.vm
<h2>Content title</h2>
<p>Content text</p>
视图/ template.vm
<html>
<head>
<title></title>
</head>
<body>
<!-- Header of page -->
#include({context variable which contains "home_view"} + ".vm");
<!-- Footer of page -->
</body>
</html>
如果有人知道CakePHP,这与其模板系统类似
我该怎么做?
答案 0 :(得分:4)
基于 @Nathan Bubna 建议的最终解决方案。
项目基于 Spring MVC 模板,具有 Maven 受控资源。
Spring版本3.1.1.RELEASE
/WEB-INF/web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" 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">
<!-- The definition of the Root Spring Container shared by all Servlets and Filters -->
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/root-context.xml</param-value>
</context-param>
<!-- Creates the Spring Container shared by all Servlets and Filters -->
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
<!-- Processes application requests -->
<servlet>
<servlet-name>mainServlet</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring/appServlet/servlet-context.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>mainServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
/WEB-INF/spring/servlet/servlet-context.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<!-- DispatcherServlet Context: defines this servlet's request-processing
infrastructure -->
<!-- Enables the Spring MVC @Controller programming model -->
<annotation-driven />
<!-- Handles HTTP GET requests for /resources/** by efficiently serving
up static resources in the ${webappRoot}/resources directory -->
<resources mapping="/resources/**" location="/resources/" />
<resources location="/resources/favicon.ico" mapping="/favicon.ico"/>
<resources location="/resources/favicon.ico" mapping="/favicon.png"/>
<beans:bean id="velocityConfig"
class="org.springframework.web.servlet.view.velocity.VelocityConfigurer">
<beans:property name="resourceLoaderPath" value="/WEB-INF/views/" />
</beans:bean>
<!-- View resolvers can also be configured with ResourceBundles or XML files.
If you need different view resolving based on Locale, you have to use the
resource bundle resolver. -->
<beans:bean id="viewResolver"
class="org.springframework.web.servlet.view.velocity.VelocityLayoutViewResolver">
<beans:property name="cache" value="true" />
<beans:property name="prefix" value="" />
<beans:property name="layoutUrl" value="layout.vm"></beans:property>
<beans:property name="suffix" value=".vm" />
</beans:bean>
<context:component-scan base-package="com.mypackage.subpackage" />
</beans:beans>
重要的pom.xml行
<dependency>
<groupId>org.apache.velocity</groupId>
<artifactId>velocity</artifactId>
<version>1.7</version>
</dependency>
<dependency>
<groupId>org.apache.velocity</groupId>
<artifactId>velocity-tools</artifactId>
<version>2.0</version>
</dependency>
答案 1 :(得分:2)
我认为Spring支持VelocityLayoutView。或者,我认为他们做到了。