Question

我正在写一个类似愤怒的小鸟的游戏，我希望当玩家输入＆＃39;然后＆＃39; o＆＃39;然后＆＃39; ; d＆＃39; （＆＃39;上帝＆＃39;，对于＆＃39;上帝模式＆＃39;），目标将在鸟当前（给定度和初始速度）着陆的地方出现（如果你＆＃39; d开火）。 - 上帝模式在游戏中可以使用3次。

所以我尝试了类，并存储了每个两个最后一个字符： [我使用tkinter和Image，来自PIL的ImageTk]

class GodMode:

times_left = 3   # Default

def __init__(self, master, image_file):
    self.master = master
    self.target_image = Image.open(image_file) 
    self.target = ImageTk.PhotoImage(self.target_image.resize((RED_CROSS_WIDTH, RED_CROSS_HEIGHT), Image.ANTIALIAS))

def display(self, bird):
    self.times_left -= 1
    self.x = physics.distance_traveled(bird)
    self.y = TARGET_Y_LOCATION
    self.master.create_image(self.x, self.y, image = self.target)

def typing_god(self, keystroke, bird):
    '''This function is first called when "g" is typed, from another 
    function (who is binded in the "main" function)
    '''
    if keystroke is None:
        self.last_char = "g"
    else:
        if keystroke.char == "d" or keystroke.char == "D":
            if self.last_char == "o" and self.char_before_last == "g":
                if self.times_left == 3:
                    self.god_mode_start(bird)
                elif 0 < self.times_left:
                    self.god_mode_alert(bird)
                else:
                    help_over()
                    return
        else:
            self.char_before_last = self.last_char
            self.last_char = keystroke.char
    self.master.bind('<Key>', lambda event: typing_god(event, bird))

def god_mode_start(self):
    start_input = messagebox.askyesno(title = "GOD Mode", message = "The god mode is only optional 3 times. Are you sure?")
    if not start_input:
        return
    self.display(bird)

def god_mode_alert(self, bird):
    start = messagebox.showinfinfo(title = "GOD Mode", message = GOD_MODE_FORMAT % self.gode_mode_left)
    if not start:
        return
    self.display(bird)

@staticmethod
def help_over():
    messagebox.showinfo(title = "GOD Mode not available", message = "You have used all your help.")

......但那不起作用，我不知道为什么。

非常感谢您的帮助！

大卫

Answer 1

看起来self.master.bind('<Key>', lambda event: typing_god(event, bird))可能永远不会被调用，你是否在这个语句周围添加了一个打印来检查？

您可能希望将其放在构造函数中，并使用self.typing_god：

def __init__(self, master, image_file):
    self.master = master
    self.target_image = Image.open(image_file) 
    self.target = ImageTk.PhotoImage(self.target_image.resize((RED_CROSS_WIDTH, RED_CROSS_HEIGHT), Image.ANTIALIAS))
    # Add a debug statement to check that this method is getting called
    print "Binding key events to self.typing_god"
    # Bind the key down event in the constructor
    self.master.bind('<Key>', lambda event: self.typing_god(event, bird))

Answer 2

执行绑定时，您可以将事件串在一起。所以，例如，你可以绑定三个字符序列＆＃34; god＆＃34; （例如："<g><o><d>"）

def god_mode(event):
    print("god mode is activated")
some_widget.bind("<g><o><d>", god_mode)

假设some_widget具有键盘焦点，当您按下按键＆＃34; g＆＃34;，＆＃34; o＆＃34;和＆＃34; d＆＃34;按此顺序，将打印消息。

这是使用虚拟活动的好机会。您可以定义自己的事件并将其绑定到它：

some_widget.event_add("<<GodMode>>", "<g><o><d>")
some_widget.bind("<<GodMode>>", god_mode)

Answer 3

我会使用deque。您可以设置maxlen = 3。当你获得新的按键时，你可以附加它。检查god中是否有deque也很容易：

In [1]: from collections import deque
In [2]: d = deque(maxlen=3)
In [3]: d.append('g') # new keystroke
In [4]: d.append('o') # new keystroke
In [5]: d.append('d') # new keystroke
In [6]: d
Out[6]: deque(['g', 'o', 'd'], maxlen=3)
In [7]: ''.join(d) == 'god'  # check if we already in the god mode
Out[7]: True
In [8]: d.append('x') # new keystroke
In [9]: d
Out[9]: deque(['o', 'd', 'x'], maxlen=3)
In [10]: ''.join(d) == 'god' # check if we already in the god mode
Out[10]: False

如何知道输入特定字符序列的时间？

3 个答案: