Flex 4 - 當狀態依賴於容器的子集時，如何循環使用容器

Question

Flex的新狀態會重新標記包含includeIn/excludeFrom的可視項目。如果我有一個Group（MainGroup）擁有5個受控制的子/元素，那麼仍然有辦法獲得對MainGroup的子項的引用嗎？ mainGroup.numChildren和mainGroup.numElements不起作用，因爲子對象已重新生成父對象。充其量，他們顯示1.

<s:states> 
    <s:State name="view1State" /> 
    <s:State name="view2State" /> 
    <s:State name="view3State" /> 
    <s:State name="view4State" /> 
</s:states> 

<s:Group id="mainGroup"> 
    <shipping:OrderShipping id="view1" 
      includeIn="view1State" /> 
    <payment:OrderPayment id="view2" 
      includeIn="view2State" /> 
    <verification:OrderVerification id="view3" 
      includeIn="view3State" /> 
    <confirmation:OrderConfirmation id="view4" 
      includeIn="view4State" /> 
</s:Group> 

Answer 1

我得到了最好的答案！那麼，我會讓你決定。只需在狀態數組上操作...

for each (var viewState:State in mainDocument.states) 
{ 
    var overrides:Array = viewState.overrides; 

    for (var i:int = 0; i < overrides.length; i++) 
    { 
     var addItems:AddItems = overrides[i]; 
     if (addItems.destination === mainDocument.mainGroup) 
      trace((addItems.items as UIComponent).name); 
    } 

} 

Answer 2

我不確定底層的狀態實現是否刪除/添加子項到組件或更改其可見性。

如果是前者，我想嘗試這樣的事：

for (var x : int = 0; x<mainGroup.numChildren ; x++){ 
var child : UIComponent = mainGroup.getChildAt(x); 
if(child.parent){ 
    // The child has a parent, so it must be displayed; do your processing 
} else { 
    // really this should never occur 
} 
} 

如果是知名度的問題，那麼這樣做：

for (var x : int = 0; x<mainGroup.numChildren ; x++){ 
var child : UIComponent = mainGroup.getChildAt(x); 
if(child.visible){ 
    // The child is displayed; do your processing 
} else { 
    // child isn't displayed 
} 
} 

我猜第一種方法是正確的一個。

[注我在瀏覽器中寫了這個代碼]

Answer 3

如果你想狀態來控制可見光/ includeinlayout而不是父母，只是做了這樣的

<s:states> 
    <s:State name="view1State" /> 
    <s:State name="view2State" /> 
    <s:State name="view3State" /> 
    <s:State name="view4State" /> 
</s:states> 
<s:Group id="mainGroup"> 
    <shipping:OrderShipping id="view1" 
     visible="false" 
     includeInLayout="false" 
     visible.view1State="true" 
     includeInLayout.view1State="true"/> 
    <payment:OrderPayment id="view2" 
     visible="false" 
     includeInLayout="false" 
     visible.view2State="true" 
     includeInLayout.view2State="true"/> 
    <verification:OrderVerification id="view3" 
     visible="false" 
     includeInLayout="false" 
     visible.view3State="true" 
     includeInLayout.view3State="true" /> 
    <confirmation:OrderConfirmation id="view4" 
     visible="false" 
     includeInLayout="false" 
     visible.view4State="true" 
     includeInLayout.view4State="true"/> 
</s:Group> 

Answer 4

我有一個類似的情況我的應用程序，這就是我的成就。我覺得有必要說，雖然這有效，但我覺得這有點不好意思。有鑑於此：

1. 創建另一個狀態，我們將其稱爲cheating。使這種狀態下，默認的（把它先在你states陣列。
2. 更新你所有的意見，包括在此狀態。
   例如，<shipping:OrderShipping id="view1" includeIn="cheating,view1State" />
3. 創建一個名爲viewsArray類型的新類成員。
4. 與引用的狀態添加creationComplete事件處理程序在處理程序populdate views到每個視圖例如，views = [view1, view2, view3, view4];
5. 更改爲你的「第一次」（非cheating）狀態：。setCurrentState("view1State");

畢竟這是你的views陣列將有每個視圖引用無論哪個國家你的應用是目前英寸