Question

4

pygi how to toggle button without running function

rated 0 times [ 4] [ 0] / answers: 1 / hits: 1503 / 1 Year ago, thu, march 23, 2023, 11:48:21

I'm writing this application using quickly.

I'm looking for a way to toggle button without running function which is connected to this button.

def on_button_text_italic_toggled(self, widget):

    print "Italic"



def on_buttone_test_clicked(self, widget):

    self.button_text_italic.set_active(True)

So I need this to make button_text_italic to be toggled but without printing out "Italic" text.

Thanks for any help!

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irripri

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answered 1 Year ago ticeate · Accepted Answer

If you want the function to be run most of the time the toggle signal is emitted, and not when you manually toggle it (e.g. when loading saved settings and displaying the appropriate state), then you need to block and unblock the signal. To do so you'll need the handle_id that was returned when the signal was connected to the function. Just assign a variable when you connect the signal. Here's an example:

#!/usr/bin/python

from gi.repository import Gtk



def on_toggle(widget,data=None):

    print "toggled, emitted signal"



def on_button1_clicked(widget, data=None):

    print "manually toggle, no signal"

    toggle.handler_block(handle_id)

    state=toggle.get_active()

    toggle.set_active(not state)

    toggle.handler_unblock(handle_id)





win=Gtk.Window()

win.connect('destroy', Gtk.main_quit)

box=Gtk.Box(orientation=Gtk.Orientation.VERTICAL)

button1=Gtk.Button('Toggle with no signal')

button1.connect('clicked', on_button1_clicked)

button1.show()

box.pack_start(button1,True,True,10)

toggle=Gtk.ToggleButton('Toggle')

handle_id=toggle.connect('toggled', on_toggle)

toggle.show()

box.pack_start(toggle,True,True,0)

box.show_all()

win.add(box)

win.show()

Gtk.main()