我们在ui:include中包含了一个页面。当点击一个按钮时,我们想要重新呈现ui:包含的页面。这是父页面;
<h:form id="mainForm">
<a4j:region rendered="#{dealHome.editMode}">
<ui:include src="/createEditDeal.xhtml">
<ui:param name="id" value="#{dealHome.instance.id}" />
</ui:include>
</a4j:region>
<a4j:region rendered="#{not dealHome.editMode}">
<ui:include src="/readDeal.xhtml">
<ui:param name="id" value="#{dealHome.instance.id}" />
</ui:include>
</a4j:region>
<s:div rendered="#{dealHome.editMode}">
<a4j:commandButton action="#{dealHome.persist()}" value="+"
id="btnTagAdd" reRender="mainForm" />
</s:div>
<s:div rendered="#{not dealHome.editMode}">
<a4j:commandButton id="btnModeChange" value="Edit"
action="#{dealHome.changeMode()}" reRender="mainForm" />
</s:div>
</h:form>
任何帮助或建议都将不胜感激。