问题
我写了一个小程序来实现秒表。此秒表将在按下时开始s
运行,并在按下时停止运行l
。为此,我使用了以下代码:
f = self.frame
w = self.window
info = Label(f,text="\nPress \'s\' to start running and \'l\' to stop running\n")
info.pack()
w.bind('<KeyPress-s>',self.startrunning)
w.bind('<KeyPress-l>',self.stoprunning)
stoprunning 和 start running 函数如下:
def startrunning(self):
r = Frame(self.window)
r.pack()
self.start = time.time()
start = Label(r,text="\nStarted running")
start.pack()
def stoprunning(self):
r = Frame(self.window)
r.pack()
self.stop = time.time()
self.timeConsumed = self.stop - self.start
Label(r,text='\nstopped running').pack()
end = Label(r,text="\nTime consumed is: %0.2f seconds" %self.timeConsumed)
end.pack(side = "bottom")
错误
按下s
键时,我收到以下错误:
>>>
Exception in Tkinter callback
Traceback (most recent call last):
File "C:\Python25\lib\lib-tk\Tkinter.py", line 1414, in __call__
return self.func(*args)
TypeError: startrunning() takes exactly 1 argument (2 given)
规格 Python 2.7
我是 tkinter 编程新手,无法理解显示此错误的内容或原因。请告诉我我是否正确使用了代码。另外请帮我解决这个问题。