在Dijit AccordionContainer中编程打开窗格的最佳方式是什么？ [英] Whats the best way to programatically open a pane inside Dijit AccordionContainer

问题描述

我正在尝试开放&以手风琴格式编程。这是我的代码的简化版本。即使我将所选的第一个窗格设置为false，将第二个窗格设置为true，只有在浏览器（FF3）上加载时，才会打开第一个窗格。

  var accordionContainer = new dijit.layout.AccordionContainer（）。placeAt（test）; 
 var accordPane = new dijit.layout.ContentPane（{title：test，content：hello}）; 
 var accordPane2 = new dijit.layout.ContentPane（{title：test1，content：hello1}）; 
 accordionContainer.addChild（accordPane）; 
 accordionContainer.addChild（accordPane2,1）; 
 accordPane.startup（）; 
 accordPane2.startup（）; 
 //accordionContainer.selectChild（accordPane2）; 
 accordionContainer.startup（）; 
 accordPane.selected = false; 
 accordPane2.selected = true; 
  

解决方案

你可以这样做：

  accordionContainer.selectChild（accordPane2）; 
  

假设您正在使用dojo 1.3。

dijit.layout.AccordionContainer 是 dijit.layout.StackContainer 的子类，它具有 selectChild 定义。

我设置了一个演示页面，您可以在此处看到此代码

如果您在之前调用 selectChild c $ c> startup ，这可能会导致您看到的错误，因为窗口小部件不处于完整状态。 （对不起，在我发布原始答案之前错过了输入的代码）

I am trying open & close accordion panes programatically. Here is the simplified version of my code. Even though I set the first pane's selected to false and and second pane's selected to true, only the first pane opens when it loads on the browser (FF3).

var accordionContainer = new dijit.layout.AccordionContainer().placeAt("test");
var accordPane = new dijit.layout.ContentPane({"title": "test", "content":"hello"});
var accordPane2 = new dijit.layout.ContentPane({"title": "test1", "content":"hello1"});
accordionContainer.addChild(accordPane);
accordionContainer.addChild(accordPane2, 1);
accordPane.startup();
accordPane2.startup();
//accordionContainer.selectChild(accordPane2);
accordionContainer.startup();
accordPane.selected = false;
accordPane2.selected = true;

解决方案

You can do it like this:

accordionContainer.selectChild( accordPane2 );

Assuming you are using dojo 1.3.

dijit.layout.AccordionContainer is a subclass of dijit.layout.StackContainer, which has selectChild defined.

I set up a demo page where you can see this code in action

If you were calling selectChild before startup, that could cause the error you were seeing since the widget wasn't in a 'complete' state. (Sorry, missed the commneted out code before I posted original answer)

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