> would it be advisable to guard against this with something like this?
>
> def perform_longrunning_calculation():
> if not app.busy:
> app.busy = 1
[...]
By using that kind of construct, instead of using update_idletasks(),
you force all code to be aware of and manage the app.busy
[EMAIL PROTECTED] wrote:
> Here's one case where it's bad to call update.
>
> def perform_longrunning_calculation():
> time.sleep(1)
> app.update()
> time.sleep(1)
>
would it be advisable to guard against this with something like this?
def perform
Here's one case where it's bad to call update.
def perform_longrunning_calculation():
time.sleep(1)
app.update()
time.sleep(1)
suppose you kick this off with a keybinding, such as:
app.bind("c", lambda e: perform_longrunning_calculat
Fredrik Lundh wrote:
> Monu Agrawal wrote:
>
>> Hi I am making a gui based tool. When user preses a perticular button I
>> am running a heavy command, before this I want to say user to wait with
>> a image showing infront of her.
>>
>> My code is like:
>>
>> def loadData(self):
>>top=Toplevel(
Monu Agrawal wrote:
> Hi I am making a gui based tool. When user preses a perticular button I
> am running a heavy command, before this I want to say user to wait with
> a image showing infront of her.
>
> My code is like:
>
> def loadData(self):
>top=Toplevel(self.parent)
>top.focus_set()
Hi I am making a gui based tool. When user preses a perticular button I
am running a heavy command, before this I want to say user to wait with
a image showing infront of her.
My code is like:
def loadData(self):
top=Toplevel(self.parent)
top.focus_set()
