2013-05-20 21 views

回答

2

我在這裏假設你想以編程方式做到這一點。看看這個答案 - https://stackoverflow.com/a/1190455/1047998 - 它描述了用於選擇特定手風琴窗格的selectChild的用法。您也可以閱讀dijit.layout.AccordionContainer的Dojo API文檔 - http://dojotoolkit.org/api/1.6/dijit/layout/AccordionContainer - 您可以在此處參考文檔selectChild

更新:

所以我們可以說,如果你確定你的可摺疊容器是這樣的:

<xp:div dojoType="dijit.layout.AccordionContainer" id="accordionContainer"> 
    <xp:div dojoType="dijit.layout.ContentPane" id="pane1" title="Pane 1"> 
     Content 1 
    </xp:div> 
    <xp:div dojoType="dijit.layout.ContentPane" title="Pane 2" id="pane2"> 
     Content 2 
    </xp:div> 
    <xp:div dojoType="dijit.layout.ContentPane" title="Pane 3" id="pane3"> 
     Content 3 
    </xp:div> 
    <xp:div dojoType="dijit.layout.ContentPane" title="Pane 4" id="pane4"> 
     Content 4 
    </xp:div> 
</xp:div> 

所以選擇pane3的JavaScript代碼將是這樣的:

var ac = dijit.byId("#{id:accordionContainer}"); 
ac.selectChild(dijit.byId("#{id:pane3}")); 
+1

這正是我需要什麼。謝謝! – Bitwyse1