I basically have a QMainWindow, and a Dialog window whose constructor is shown below;
class VisualTool(QtGui.QDialog):
def __init__(self, parent = None):
QtGui.QWidget.__init__(self, parent)
self.WidgetBoard = Ui_Aesthetics_Tool()
self.WidgetBoard.setupUi(self)
self.setWindowFlags(QtCore.Qt.Tool) # <-
As you can see, I would like to treat the Dialog as a Tool window (it’s exactly the type of window I need). The tool window should be shown after a button click on the QMainWindow, and for interaction with QMainWindow to continue.
Before my QMainWindow is shown, calling .show() on my tool window spawns it correctly.
However, if I attempt to show my Tool window AFTER showing QMainWindow (such as after a button click), calling .show() and .exec() have no effect whatsoever.
(There’s not even any flicker of a window. There’s no spawn whatsoever!)
Once the window is shown, I can not change the Window Flags. It has no effect.
How can I get this Tool window to show?!
Thanks!
PyQt4
python 2.7.2
windows 7
I’m not sure I understand what your issue really is. This test code snippet seems to function fine, though I have no idea what your missing Ui_Aesthetics_Tool() code does to modify the tool window:
I am able to launch multiple tool windows.
And as for using exec_(), that is a modal blocking call and probably not what you want.