On Tue, 22 Apr 2008 21:31:35 -0700
John Finlay [EMAIL PROTECTED] wrote:
It does inherit the visible property - what makes you think it doesn't?
The exception that my app threw when I tried to use it:
Traceback (most recent call last):
File /monitor/track.py, line 131, in start_updates
if
2008/4/23, Mitko Haralanov [EMAIL PROTECTED]:
The exception that my app threw when I tried to use it:
Traceback (most recent call last):
File /monitor/track.py, line 131, in start_updates
if syslog_frame.visible:
AttributeError: 'gtk.Frame' object has no attribute 'visible'
If you
Mitko Haralanov wrote:
On Tue, 22 Apr 2008 21:31:35 -0700
John Finlay [EMAIL PROTECTED] wrote:
It does inherit the visible property - what makes you think it doesn't?
The exception that my app threw when I tried to use it:
Traceback (most recent call last):
File /monitor/track.py, line
On Wed, 23 Apr 2008 14:10:10 -0300
Facundo Batista [EMAIL PROTECTED] wrote:
If you could provide an example that makes this to you, for us to try
it, we could help you better.
Here you go:
#!/usr/bin/python
import gtk
import pygtk
def toggle_func (widget, *user_data):
frame = user_data[0]
On Wed, 23 Apr 2008 13:23:22 -0400
John Ehresman [EMAIL PROTECTED] wrote:
Try
syslog_frame.props.visible
That worked. Thank you.
--
Mitko Haralanov
==
43. If I knew it wasn't going to work, I would have tested it sooner.
--Top 100 things
What is the best way to get whether a widget is visible or not?
According to the documentation, gtk.Widget has the 'visible' property
which is Read-Write. However, gtk.Frame for example does not seem to
inherit that property of gtk.Widget.
Thanks for the help!
--
Mitko Haralanov
Mitko Haralanov wrote:
What is the best way to get whether a widget is visible or not?
According to the documentation, gtk.Widget has the 'visible' property
which is Read-Write. However, gtk.Frame for example does not seem to
inherit that property of gtk.Widget.
It does inherit the visible
