spring roo excel view

Question

我试图输出一个Excel文件作为Spring Roo 1.0.2中的视图.最快的方法是什么？(我是否必须添加新的映射等？)目前我正在使用默认的Roo AjaxUrlBasedViewResolver.

谢谢

Answer 1

我不认为有一种特定的"Roo"方式,但是Spring MVC确实有一个使用Apache POI的AbstractExcelView类,并且不会引起任何问题 - 我认为这是你能想到的最好的.

首先创建一个扩展AbstractExcelView并实现buildExcelDocument方法的视图类:

 import org.springframework.web.servlet.view.document.AbstractExcelView;

 public class XlsView  extends AbstractExcelView { 

   @Override
   protected void buildExcelDocument(
            Map<String, Object> model, HSSFWorkbook workbook, 
            HttpServletRequest request, HttpServletResponse response) throws Exception {
      //Apache POI code to set up the HSSFWorkbook goes here
      response.setHeader("Content-Disposition", "attachment; filename=\"file.xls\"");
    }
 }

然后将以下内容添加到webmvc-config.xml中.我不认为这个位置很重要,但我在下面列出了我的TilesConfigurer:

   <bean id="excelViewResolver" class="org.springframework.web.servlet.view.XmlViewResolver">
            <property name="order" value="1"/>
            <property name="location" value="/WEB-INF/views/views-excel.xml"/>
        </bean>

最后创建一个views-excel.xml,其中包含以下内容:

  <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xsi:schemaLocation="http://www.springframework.org/schema/aop   http://www.springframework.org/schema/aop/spring-aop-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   http://www.springframework.org/schema/jee http://www.springframework.org/schema/jee/spring-jee-3.0.xsd   http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.0.xsd">
     <bean name="XlsView" class="com.your.package.XlsView"/>
  </beans>