스케일러 템플릿이 작동하지 않습니다.

Question

커플 일 전 스케이드 템플릿 endine을 사용하여 스프링 webapp를 만들었습니다. 모든 것이 훌륭하게 작동했지만 어제부터 나는 그것을 실행할 수 없습니다. 이것은 내가

org.springframework.web.util.NestedServletException: Handler processing failed; nested exception is java.lang.VerifyError: (class: org/fusesource/scalate/spring/view/ScalateUrlView, method: renderMergedTemplateModel signature: (Ljava/util/Map;Ljavax/servlet/http/HttpServletRequest;Ljavax/servlet/http/HttpServletResponse;)V) Incompatible argument to function 
org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:839) 
org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:719) 
org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:669) 
org.springframework.web.servlet.FrameworkServlet.doGet(FrameworkServlet.java:574) 
javax.servlet.http.HttpServlet.service(HttpServlet.java:617) 
javax.servlet.http.HttpServlet.service(HttpServlet.java:717) 

root cause 

java.lang.VerifyError: (class: org/fusesource/scalate/spring/view/ScalateUrlView, method: renderMergedTemplateModel signature: (Ljava/util/Map;Ljavax/servlet/http/HttpServletRequest;Ljavax/servlet/http/HttpServletResponse;)V) Incompatible argument to function 
org.fusesource.scalate.spring.view.ScalateViewResolver.buildView(ScalateViewResolver.scala:41) 
org.springframework.web.servlet.view.UrlBasedViewResolver.loadView(UrlBasedViewResolver.java:419) 
org.springframework.web.servlet.view.AbstractCachingViewResolver.createView(AbstractCachingViewResolver.java:158) 
org.springframework.web.servlet.view.UrlBasedViewResolver.createView(UrlBasedViewResolver.java:384) 
org.springframework.web.servlet.view.AbstractCachingViewResolver.resolveViewName(AbstractCachingViewResolver.java:77) 
org.springframework.web.servlet.DispatcherServlet.resolveViewName(DispatcherServlet.java:1078) 
org.springframework.web.servlet.DispatcherServlet.render(DispatcherServlet.java:1027) 
org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:817) 
org.springframework.web.servlet.DispatcherServlet.doService(DispatcherServlet.java:719) 
org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:669) 
org.springframework.web.servlet.FrameworkServlet.doGet(FrameworkServlet.java:574) 
javax.servlet.http.HttpServlet.service(HttpServlet.java:617) 
javax.servlet.http.HttpServlet.service(HttpServlet.java:717) 

내가하지

이 내 서블릿의 context.xml

이다가 실행되지만 scalate에 나는 할 수있는 JSP 내 서블릿-context.xml에 다시 전환 할 수 있습니다 때 무엇을 얻을

<?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"> 

<!-- 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/" /> 

<beans:bean class="org.fusesource.scalate.spring.view.ScalateViewResolver"> 
    <beans:property name="prefix" value="/WEB-INF/scalate/" /> 
    <beans:property name="suffix" value=".ssp" /> 
</beans:bean> 

<context:component-scan base-package="org.fixus.springer" /> 

내가 문제를 해결했습니다

Answer 1

도와주세요. 나는 pom.xml scalate-core에서 삭제하고 문제를 해결했습니다.