[Spring] 스프링 servlet-context로 jsp 경로 설정하기

 

스프링에서 빈을 이용한 jsp 경로 설정 방법입니다.

 

<?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 https://www.springframework.org/schema/mvc/spring-mvc.xsd
		http://www.springframework.org/schema/beans https://www.springframework.org/schema/beans/spring-beans.xsd
		http://www.springframework.org/schema/context https://www.springframework.org/schema/context/spring-context.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/" />

	<!-- Resolves views selected for rendering by @Controllers to .jsp resources in the /WEB-INF/views directory -->
	<beans:bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
		<beans:property name="prefix" value="/WEB-INF/views/" />
		<beans:property name="suffix" value=".jsp" />
	</beans:bean>
	
	<context:component-scan base-package="com.myspring.pro27" />
	
	
	
</beans:beans>

 

bean을 이용하여 viewResolver 클래스의 속성을 위와 같이 설정해주시면 됩니다.

 

ModelAndView mav = new ModelAndView("home");

 

views로 기준으로 상대경로를 적어주시면 됩니다.

 

만약 views/list/home.jsp 라면 view의 경로는 /list/home입니다. 뒤에 jsp를 붙이시면 안됩니다.