python - 如何通过按下按钮关闭程序？

Question

我希望你能帮我解决这个问题..

我有以下代码:

from Tkinter import *
import ttk
import time

def start():
    start_stop.config(text="QUIT", command=stop)
    for i in xrange(5):
        pbar_det.step(19.99)
        master.update()
        # Busy-wait
        time.sleep(1)
    master.quit()

def stop():
    master.quit()

master = Tk()

start_stop = Button(master, text='START', command=start)
start_stop.grid(row=0, column=1, pady=2, padx=2, sticky=E+W+N+S)

pbar_det = ttk.Progressbar(master, orient="horizontal", length=600, mode="determinate")
pbar_det.grid(row=0, column=0, pady=2, padx=2, sticky=E+W+N+S)

master.mainloop()

我按下开始按钮，进度条开始，按钮文本变为“QUIT”并在进度条满(五秒)时结束(并且程序关闭)

当时我该怎么做才能按“QUIT”关闭程序？ (不等待柱填充)

希望你能帮帮我!谢谢!

Answer 1

试试这个:

from Tkinter import *
import ttk
import time
import threading

def start():
    start_stop.config(text="QUIT", command=stop)
    thread = threading.Thread(target=progBar, args=())
    thread.daemon = True
    thread.start()

def progBar():
    for i in xrange(5):
        pbar_det.step(19.99)
        master.update()
        # Busy-wait
        time.sleep(1)
    master.quit()

def stop():
    master.quit()

master = Tk()

start_stop = Button(master, text='START', command=start)
start_stop.grid(row=0, column=1, pady=2, padx=2, sticky=E+W+N+S)

pbar_det = ttk.Progressbar(master, orient="horizontal", length=600, mode="determinate")
pbar_det.grid(row=0, column=0, pady=2, padx=2, sticky=E+W+N+S)

master.mainloop()

编辑

变量名固定。