如何为特定视图定义图块定义

Question

所以我的图块定义如下所示，我想知道如何管理它以在特定视图上显示特定模板：我想要默认模板 ist "commonPage" 和例如"/monitor1" return 值我希望应用空模板。

<tiles-definitions>
  <definition name="commonPage" template="/WEB-INF/tiles/layouts/topmenulayout.jsp">
    <put-attribute name="header" value="/WEB-INF/tiles/jsp/header.jsp" />
    <put-attribute name="menu" value="/WEB-INF/views/menu.jsp" />
    <put-attribute name="footer" value="/WEB-INF/tiles/jsp/footer.jsp" />
  </definition>

  <definition name="emptyPage" template="/WEB-INF/tiles/layouts/emptylayout.jsp">
  </definition>


  <!-- Views ending in "Dialog" have no header, menu or      -->
  <!-- footer. Useful for creating jQuery popup dialogs.     -->
  <definition name="*/*Dialog" template="/WEB-INF/views/{1}/{2}Dialog.jsp" />
  <definition name="*Dialog" template="/WEB-INF/views/{1}Dialog.jsp" />

  <!-- All other views get the header, menu and footer tiles. -->
  <!-- I want commonPage as default template, and e.g. for "/Monitor1" return value I want the emptyLayout       -->

  <definition name="*" extends="commonPage">
    <put-attribute name="body" value="/WEB-INF/views/{1}.jsp" />
  </definition>
</tiles-definitions>

Answer 1

很容易解决。我添加了以下定义，它按照我想要的方式工作：

<definition name="*Monitor1" extends="emptyPage">

如何为特定视图定义图块定义

How to define tiles definition for a specific view

spring-mvc

tiles

apache-tiles