ActionScript 3的使用多的onkeydown /的onkeyup

Question

林試圖讓蛇的兩個PLYER版本上運行，但我遇到麻煩第二蛇的工作，球員1次用W，A，S和d，玩家2使用箭頭鍵。有w，a，s和d的玩家1正在工作，玩家2的箭頭沒有。

的代碼看起來是這樣的：

function onKeyDown(event:KeyboardEvent):void{ 
     if(event.keyCode == Keyboard.A){ 
      SnakeDirection = "left"; 
     }else if (event.keyCode == Keyboard.D) { 
      SnakeDirection = "right"; 
     }else if (event.keyCode == Keyboard.W) { 
      SnakeDirection = "up"; 
     }else if (event.keyCode == Keyboard.S) { 
      SnakeDirection = "down"; 
     } 
    } 
    function onKeyUp(event:KeyboardEvent):void { 
     if(event.keyCode == Keyboard.A) { 
      SnakeDirection = ""; 
     }else if(event.keyCode == Keyboard.D) { 
      SnakeDirection = ""; 
     }else if(event.keyCode == Keyboard.W) { 
      SnakeDirection = ""; 
     }else if(event.keyCode == Keyboard.S){ 
      SnakeDirection = ""; 
     } 
    } 
/*  function onKeyDown(event:KeyboardEvent):void{ 
      if(event.keyCode == Keyboard.LEFT){ 
       Snake2Direction = "left"; 
      }else if (event.keyCode == Keyboard.RIGHT) { 
       Snake2Direction = "right"; 
      }else if (event.keyCode == Keyboard.UP) { 
       Snake2Direction = "up"; 
      }else if (event.keyCode == Keyboard.DOWN) { 
       Snake2Direction = "down"; 
      } 
     } 
     function onKeyUp(event:KeyboardEvent):void { 
      if(event.keyCode == Keyboard.LEFT) { 
       Snake2Direction = ""; 
      }else if(event.keyCode == Keyboard.RIGHT) { 
       Snake2Direction = ""; 
      }else if(event.keyCode == Keyboard.UP) { 
       Snake2Direction = ""; 
      }else if(event.keyCode == Keyboard.DOWN){ 
       Snake2Direction = ""; 
      } 
     }*/ 

的意見是有，因爲它打破了比賽，從我的理解是錯誤的，因爲我只能用一個的onkeyup /向下。如果是這樣的話，還有其他的方法嗎？

謝謝！

Answer 1

你爲什麼要使用2個功能爲每個事件？如果你只是在同樣的函數中完成第二個球員的邏輯，那麼它會不會更簡單？它也處理第一個球員的邏輯？

像這樣：

function onKeyDown(event : KeyboardEvent) : void { 
     //handle player 1 
     if (event.keyCode == Keyboard.A) { 
      SnakeDirection = "left"; 
     } else if (event.keyCode == Keyboard.D) { 
      SnakeDirection = "right"; 
     } else if (event.keyCode == Keyboard.W) { 
      SnakeDirection = "up"; 
     } else if (event.keyCode == Keyboard.S) { 
      SnakeDirection = "down"; 
     } 
     //handle player 2 
     if (event.keyCode == Keyboard.LEFT) { 
      Snake2Direction = "left"; 
     } else if (event.keyCode == Keyboard.RIGHT) { 
      Snake2Direction = "right"; 
     } else if (event.keyCode == Keyboard.UP) { 
      Snake2Direction = "up"; 
     } else if (event.keyCode == Keyboard.DOWN) { 
      Snake2Direction = "down"; 
     } 
    } 

    function onKeyUp(event : KeyboardEvent) : void { 
     //handle player 1 
     if (event.keyCode == Keyboard.A || event.keyCode == Keyboard.D || event.keyCode == Keyboard.W || event.keyCode == Keyboard.S) { 
      SnakeDirection = ""; 
     } 
     //handle player 2 
     if (event.keyCode == Keyboard.LEFT ||event.keyCode == Keyboard.RIGHT || event.keyCode == Keyboard.UP || event.keyCode == Keyboard.DOWN) { 
      Snake2Direction = ""; 
     } 
    } 

請記住，你要保持如果塊用於分離的球員，這樣就可以同時獲得多個輸入。

另外，作爲一個側面說明，否則，如果在你的onKeyUp功能塊是不必要的。無論如何，他們都做同樣的事情。