You're fro loop could be speeded with a generator :
root.__set_children__( i for i in root.__get_children__() if i is
not child )
But I dont really understand the propose qof you try except. (in most
case, a bare except is avery bad idea (since it begin to catch Syntax
error and such.)
On 3 juil. 07, at 01:44, Lunpa, The wrote:
> Thats good in some cases for persistent widgets that you have a known
> quantity of at all times.
>
> I've built a work around, since the closest thing I can find is
> __get_children__() and __set_children__()
> Its a bit crude, but it works, so I'll post it:
>
> def remove_widget(child, root=None):
> """Because there is no RootWidget.remove_child(foo) method!"""
> try:
> if not root:
> root = soya.root_widget
> old_children = root.__get_children__()
> children = [ ]
> for i in old_children:
> if i != child:
> children.append(i)
>
> root.__set_children__(children)
> return True
> except:
> return False
>
>
> On 7/2/07, Holcroft Jean-Baptiste <[EMAIL PROTECTED]> wrote:
>> maybe you can hide your child
>>
>> 2007/7/2, Lunpa, The <[EMAIL PROTECTED]>:
>>> Once a child has been added to in pudding (such as with
>>> pudding.RootWidget.add_child), is there a good way to _remove_
>>> it? I
>>> can't find anything like pudding.RootWidget.remove_child() or
>>> whatnot.
>>>
>>> _______________________________________________
>>> Soya-user mailing list
>>> [email protected]
>>> https://mail.gna.org/listinfo/soya-user
>>>
>>
>>
>> _______________________________________________
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>> [email protected]
>> https://mail.gna.org/listinfo/soya-user
>>
>>
>
>
> --
> This email message is public domain. Have a nice day! ^_^
>
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marmoute
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