Flex:为viewstack创建多列导航栏

发布于 09-26 08:17 字数 3065 浏览 9 评论 0原文

我创建了一个 Viewstack 并使用 Tile 组件和重复 LinkBut​​tons 我能够使用 viewstack 作为数据提供者进行多列导航。我的问题是这可以做得更好吗?我的代码如下,我想知道我是否采用了这种方法。

<?xml version="1.0" encoding="utf-8"?>
<s:Application xmlns:fx="http://ns.adobe.com/mxml/2009"
               xmlns:s="library://ns.adobe.com/flex/spark"
               xmlns:mx="library://ns.adobe.com/flex/mx">
    <s:layout>
        <s:BasicLayout />
    </s:layout>

    <fx:Script>
        <![CDATA[
            import mx.controls.Alert;

            private var _listItem:Object;
            private var n:int=0;
            public function get listItem():Object
            {
                return this._listItem;
            }

            public function set listItem(listItem:Object):void
            {
                try{n++;
                    this.changeSelection(this._listItem);               
                }catch(e:Error){}
                if(n==1 || n > this.viewStack.length){
                    this._listItem = listItem;
                    this.changeSelection(listItem);
                }
            }

            private function setSelection(obj:Object):void{
                this.viewStack.selectedIndex = this.viewStack.getChildIndex(this.viewStack.getChildByName(obj.target.getRepeaterItem().name));
                this.listItem = obj.target; 
            }
            private function checkSelection(obj:Object):void{               
                if(obj.target.getRepeaterItem() == this.viewStack.selectedChild){
                    if(this.listItem != obj.target){
                        this.listItem = obj.target; 
                    }
                }               
            }
            private function changeSelection(obj:Object):void{              
                if(obj.getRepeaterItem() == this.viewStack.selectedChild){
                    obj.setStyle("color","#000000");    
                }else{
                    obj.setStyle("color","#999999");
                }               
            }
        ]]>
    </fx:Script>

    <mx:Tile id="tiles" horizontalGap="20" verticalGap="0" y="210" direction="vertical">        
        <mx:Repeater id="masterList" dataProvider="{viewStack}">
            <mx:LinkButton 
                id="btn" 
                label="{masterList.currentItem.label}" 
                click="this.setSelection(event)"
                color="#999999"
                creationComplete="checkSelection(event);" />
        </mx:Repeater>
    </mx:Tile>

    <mx:ViewStack id="viewStack"  height="200" width="300" backgroundColor="#000000" >      
        <mx:VBox id="vb1" backgroundColor="#FF0000" label="Screen One"/>        
        <mx:VBox id="vb2" backgroundColor="#00FF00" label="Screen Two"/>        
        <mx:VBox id="vb3" backgroundColor="#0000FF" label="Screen Three"/>          
        <mx:VBox id="vb4" backgroundColor="#00FFFF" label="Screen Four"/>  
    </mx:ViewStack>

</s:Application>

I have created a Viewstack and using a Tile component and repeating LinkButtons I was able to make a multi column navigation with the viewstack as the dataprovider. My question is can this be done better? My code is below and I am wondering if I took the long way around this approach.

<?xml version="1.0" encoding="utf-8"?>
<s:Application xmlns:fx="http://ns.adobe.com/mxml/2009"
               xmlns:s="library://ns.adobe.com/flex/spark"
               xmlns:mx="library://ns.adobe.com/flex/mx">
    <s:layout>
        <s:BasicLayout />
    </s:layout>

    <fx:Script>
        <![CDATA[
            import mx.controls.Alert;

            private var _listItem:Object;
            private var n:int=0;
            public function get listItem():Object
            {
                return this._listItem;
            }

            public function set listItem(listItem:Object):void
            {
                try{n++;
                    this.changeSelection(this._listItem);               
                }catch(e:Error){}
                if(n==1 || n > this.viewStack.length){
                    this._listItem = listItem;
                    this.changeSelection(listItem);
                }
            }

            private function setSelection(obj:Object):void{
                this.viewStack.selectedIndex = this.viewStack.getChildIndex(this.viewStack.getChildByName(obj.target.getRepeaterItem().name));
                this.listItem = obj.target; 
            }
            private function checkSelection(obj:Object):void{               
                if(obj.target.getRepeaterItem() == this.viewStack.selectedChild){
                    if(this.listItem != obj.target){
                        this.listItem = obj.target; 
                    }
                }               
            }
            private function changeSelection(obj:Object):void{              
                if(obj.getRepeaterItem() == this.viewStack.selectedChild){
                    obj.setStyle("color","#000000");    
                }else{
                    obj.setStyle("color","#999999");
                }               
            }
        ]]>
    </fx:Script>

    <mx:Tile id="tiles" horizontalGap="20" verticalGap="0" y="210" direction="vertical">        
        <mx:Repeater id="masterList" dataProvider="{viewStack}">
            <mx:LinkButton 
                id="btn" 
                label="{masterList.currentItem.label}" 
                click="this.setSelection(event)"
                color="#999999"
                creationComplete="checkSelection(event);" />
        </mx:Repeater>
    </mx:Tile>

    <mx:ViewStack id="viewStack"  height="200" width="300" backgroundColor="#000000" >      
        <mx:VBox id="vb1" backgroundColor="#FF0000" label="Screen One"/>        
        <mx:VBox id="vb2" backgroundColor="#00FF00" label="Screen Two"/>        
        <mx:VBox id="vb3" backgroundColor="#0000FF" label="Screen Three"/>          
        <mx:VBox id="vb4" backgroundColor="#00FFFF" label="Screen Four"/>  
    </mx:ViewStack>

</s:Application>

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马蹄踏│碎落叶2024-10-03 08:17:54

在我看来,您的导航链接公开了不同的链接,并且这些链接根据选择的链接而改变颜色。假设情况确实如此,这听起来非常像基于选项卡的导航模型。我的方法是使用 Spark TabBar 并将选项卡设置为看起来像链接的外观。这样您就可以摆脱大部分代码,并让选项卡皮肤根据当前状态来处理颜色更改。另外,您不需要任何用于更改视图堆栈的代码,因为 TabBar 会为您处理该问题。希望有帮助。

Looks to me like you've got navigation links that expose different and that those links change color based on which one is selected. Assuming that's the case, it sounds an awful lot like a tab-based navigation model. My approach would be to use the spark TabBar and skin the tabs to look like links. That way you can get rid of most of your code and let the tab skin handle changing the colors based on their current state. Also, you wouldn't need any of the code you have for changing the view stack since the TabBar would handle that for you. Hope that helps.

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