huey.y.ji...@gmail.com wrote:
> Hi Folks,
>
> I need to run an infinite loop on a Menu button, like this:
>
> from Tkinter import *
>
> def run_job():
> i = 0
> while 1:
> i = i + 1
> if i > 100:
> break
> .......
>
> root = Tk()
> menu = Menu(root)
> root.config(menu=menu)
> filemenu = Menu(menu)
> menu.add_cascade(label="Jobs", menu=filemenu)
> filemenu.add_command(label="Run", command=run_job)
> mainloop()
>
>
> In fact, the def run_job has more things to do. But, it is an infinite
> loop in essence. It works. However, when the def run_job is running, the
> menu will be hang there. I have no way to stop it. Even if I raise it with
> Key-Interupt, the thing was the same: the infinite loop cannot be stopped
> nicely. The only way was to wait until the Python interpreter notice the
> problem, and popped out an error message, then I can kill the running
> program.
>
> Can anybody tell me how to handle this? I wish when variable i reached
> 100, the run_job will be end, and the Menu "Run" will be free.
>
> Thanks in advance.
You need to break your loop into smaller steps and after each step give the
user interface a chance to run.
Here's a fancy way using a generator. run_job() implements the steps and
after each yield next_step() will schedule itself for reinvocation after 100
milliseconds.
from Tkinter import *
def run_job():
i = 0
while True:
status("i = {}".format(i))
i = i + 1
if i > 100:
break
yield
def start_job():
global steps
status("starting")
filemenu.entryconfig("Run", state=DISABLED)
steps = run_job()
next_step()
def next_step():
try:
next(steps)
except StopIteration:
status("done")
filemenu.entryconfig("Run", state=NORMAL)
else:
root.after(100, next_step)
def status(message):
status_label["text"] = message
print(message)
root = Tk()
menu = Menu(root)
root.config(menu=menu)
root.minsize(100, 50)
filemenu = Menu(menu)
menu.add_cascade(label="Jobs", menu=filemenu)
filemenu.add_command(label="Run", command=start_job)
status_label = Label(root)
status_label.pack()
root.mainloop()
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