动作脚本3,角色选择

不要使用布尔值,而是使用一个属性来存储您选择的字符:

private var player:Player;

选择角色时,只需更新该属性:

player = timmy; // or caroline

这样,您的所有代码都只需要引用它,并且可以更加通用:

private function movement():void 
{
    if(player === null)
    {
        // Don't run this function if neither character has been selected yet.
        return;
    }

    if(rightKey) player.moveRight();
    if(leftKey) player.moveLeft();      
    if(upKey) player.moveUp();
}

您正在寻找的是所谓的状态机或状态设计模式。因此,无论选择什么角色,他们都使用类似的方法: 向左移动() , 向右移动() 和 上移() .

“四人帮”论国家设计模式:

允许对象在其内部状态更改时更改其行为。对象将显示为更改其类。

因此,您所提供的代码的一个基本思想是对两个字符使用相同的状态机,并从该状态机中操作正确的对象。以下是如何做到这一点的示例:

创建状态机!:

import flash.ui.Keyboard;

class StateMachine {
    private var character:Player;
    private var state:String; //store the current state we are in, right, left, up etc

    public function StateMachine( character:Player ) {
        this.character = character; //set the character to control states on
    }

    //handle the input from the keyboard event
    public function handleInput( event:Object ):void {
        switch( event.type ) {
            case "keyDown":
                if ( event.keyCode == Keyboard.RIGHT ) {
                    state = Keyboard.RIGHT;
                }
                if ( event.keyCode == Keyboard.UP ) {
                    state = Keyboard.UP;
                }
                if ( event.keyCode == Keyboard.LEFT ) {
                    state = Keyboard.LEFT;
                }
                if ( event.keyCode == Keyboard.DOWN ) {
                    state = Keyboard.DOWN;
                }
                break;
            case "keyUp":
                if ( event.keyCode == state ) {
                    state = 0;
                }
                break;
        }
    }

    //on our update, apply the right function
    public function update():void {
        if ( state == KEYBOARD.RIGHT ) {
            character.moveRight(); 
        }
        if ( state == KEYBOARD.DOWN ) {
            character.moveDown(); 
        }
        if ( state == KEYBOARD.LEFT) {
            character.moveLeft(); 
        }
        if ( state == KEYBOARD.UP ) {
            character.moveUp(); 
        }
    }

现在,在处理角色移动时,最好的部分是让状态机处理,而不是为正确的玩家使用所有布尔值。因此,当选择角色时,我们可以创建一个新的状态机:

private var stateMachine:StateMachine;

private function keyUp(e:KeyboardEvent):void 
{

    if (e.keyCode == 38)
    {
        if (playPrincess)
        {
            upKey = false;
            stateMachine = new StateMachine( caroline );
            caroline.standStill();
        }
        else
        if (playSirTimmy)
        {
            upKey = false;
            stateMachine = new StateMachine( timmy ); 
            timmy.standStill();
        }
    }
}

现在,您可以只使用一个移动功能来处理键盘输入 KEY_DOWN 和 KEY_UP 此函数的事件,状态机处理这两个事件:

private function onKeyMovement( e:KeyboardEvent ):void {
   stateMachine.handleInput( e.keyCode );
}

最后,在更新函数中(无论是事件帧还是类似的内容),调用 stateMachine.update(); 即:

private function myUpdateFunction( e:Event ):void {
    stateMachine.update();
}

这应该会删除所有键布尔值,并让状态机处理所有输入。

希望这有帮助!