Flex 菜单控件 - 单击一个按钮并显示一个菜单.如何再次单击该按钮并隐藏该菜单? [英] Flex Menu Control - Click a button and a menu is displayed. How can I click that button a second time and hide that menu?

问题描述

基本上，我有一个按钮，点击它会显示一个菜单.我想再次单击该菜单并关闭菜单.目前，每次单击该按钮时，菜单都会重新打开.我粘贴了下面的 Flex livedoc 示例.如果您单击该按钮，菜单会不断重新打开.

Basically, I have a button and on click it displays a menu. I want to click that menu a second time and the menu closes. Currently, every time you click the button, the menu reopens. I pasted the Flex livedoc example below. If you click the button, the menu keeps reopening.

现在，我通过设置一个变量来打开和关闭它，所以当点击按钮时它会进行检查.但是，如果您在远离屏幕的地方单击，则会调度 HIDE 事件并关闭菜单.这弄乱了设置的打开关闭变量.

Now, I rigged it up by setting a var to open and closed, so when clicking the button it does a check. However, if you click away from the screen, the HIDE event gets dispatched, and the menu closes. This messed up the open close var being set.

如何让下面的 Flex 示例在单击按钮时显示菜单，然后在第二次单击按钮时关闭菜单?考虑到如果您从菜单中点击，它会关闭它.

How could I make this Flex example below show the menu on button click, and then on a second button click, it closes the menu? Take into affect that if you click away from the menu, it closes it.

此外，我使用了按钮的 MOUSE_DOWN_OUTSIDE 事件并设置了 preventDefault，并将 FlexMouseEvent event.cancelable 设置为 false.

Also, I played around with the MOUSE_DOWN_OUTSIDE event for the button and set the preventDefault, and the FlexMouseEvent event.cancelable is set to false.

更改为 PopUpMenuButton 不是一种选择.我需要涉及很多皮肤.

Changing to a PopUpMenuButton is not an option. I have to much skinning involved.

这是 Flex 示例:

Here is the Flex example:

<mx:Script>
    <![CDATA[
        // Import the Menu control.
        import mx.controls.Menu;

        // Create and display the Menu control.
        private function createAndShow():void {
            var myMenu:Menu = Menu.createMenu(null, myMenuData, false);
            myMenu.labelField="@label";
            myMenu.show(10, 10);
        }
    ]]>
</mx:Script>

<!-- Define the menu data. -->
<mx:XML format="e4x" id="myMenuData">
    <root>
        <menuitem label="MenuItem A" >
            <menuitem label="SubMenuItem A-1" enabled="false"/>
            <menuitem label="SubMenuItem A-2"/>
        </menuitem>
        <menuitem label="MenuItem B" type="check" toggled="true"/>
        <menuitem label="MenuItem C" type="check" toggled="false"/>
        <menuitem type="separator"/>     
        <menuitem label="MenuItem D" >
            <menuitem label="SubMenuItem D-1" type="radio" 
                groupName="one"/>
            <menuitem label="SubMenuItem D-2" type="radio" 
                groupName="one" toggled="true"/>
            <menuitem label="SubMenuItem D-3" type="radio" 
                groupName="one"/>
        </menuitem>
    </root>
</mx:XML>

<mx:VBox>
    <!-- Define a Button control to open the menu -->
    <mx:Button id="myButton" 
        label="Open Menu" 
        click="createAndShow();"/>
</mx:VBox>

推荐答案

//Declare menu as an instance variable instead of a local var
private var myMenu:Menu;

//var to store menu status
private var isMenuVisible:Boolean

//Create the Menu control. call this from the creationComplete of the 
//application or the Component that it is part of.
private function createMenu():void 
{
    var myMenu:Menu = Menu.createMenu(null, myMenuData, false);
    myMenu.labelField="@label";
    //menu fires an event when it is hidden; listen to it.
    myMenu.addEventListener(MenuEvent.MENU_HIDE, onMenuHidden);
}
private function onMenuHidden(e:MenuEvent):void
{
    /*
    menuHide event fired whenever the menu or one of its submenus
    are hidden - makes sure it was indeed the main menu that was hidden
    I don't have compiler handy to test this, so if for 
    some reason comparing myMenu with e.menu doesn't work,
    try if(e.target == myMenu) instead; 
    And please let me know which one works via comment :)
    */

    if(e.menu == myMenu)
       isMenuVisible = false;
}
//call this from button's click 
private function toggleMenu():void
{
    if(isMenuVisible)
        myMenu.hide();
    else
        myMenu.show();
}

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