Qt 5.4.0 on Windows, isActive() called on QWidget in some cases returns obviously wrong value

Question

In some cases on Windows platform, isActive() called on QWidget returns true, when, obviously, it cannot be active.

For example, my app doing much work on GUI thread and shows window with some delay after launch. If I launch app and switch to another application before app shows main window, when window will be showed, it will be definitely inactive. But calling isActive() returns true in this case. If I switch to my app and switch back to another app, isActive() will be false, as normal. But at first time it is true, which isn't what should normally be. On linux and OS X it works as planned, by the way.

How can I fix it? Or how can I workaround it?

Example code reproducing that issue given below. It's PyQt, but trust me, Qt reproduces this issue too.

import sys
from time import sleep
from PyQt5.QtWidgets import QWidget, QApplication

class BadWidget(QWidget):
    def __init__(self):
        super(BadWidget, self).__init__()
        self.startTimer(500)

    def timerEvent(self, e):
        print(self.isActiveWindow() and not self.isMinimized())

if __name__ == '__main__':
    app = QApplication(sys.argv)
    bad_widget = BadWidget()
    sleep(1)
    bad_widget.show()
    app.exec_()

On Windows, apparently, there's no way to get a valid `isActive` result until all native events have been processed. I think it's a platform shortcoming. — Kuba hasn't forgotten Monica, Mar 20 '15 at 15:22

score 2 · Accepted Answer · edited May 23 '17 at 11:43

Finally, I found some working solution which does exatly what I want. Unfortunately, it can be done only using WinAPI call, but it's OK.

The main idea is that on Windows, "active" and "foreground" window are not the same thing. More info here: https://stackoverflow.com/a/3940383/3400560

Final solution looks like this:

if sys.platform.startswith('win'):
    if ctypes.c_long(self.effectiveWinId()).value ==\
            windll.user32.GetForegroundWindow():
        # active
    else:
        # inactive
else:
    if self.isActiveWindow():
        # active
    else:
        # inactive

score 0 · Answer 2 · answered Mar 20 '15 at 13:15

0

Bad solution: call QApplication::processEvents() before requesting widget state.

Good solution: do "much work" in a thread pool via QtCuncurrent or other multi-threaded ways.

answered Mar 20 '15 at 13:15

Dmitry Sazonov

8,801
1
35
61

This method shows interesting behaviour. I change sleep(1) to for loop with sleep(0.01) and call to processEvents(). With this changes, before Windows shows QWidget, application spams "False" in console, so, window is inactive and everything is great. But when QWidget appears, app starts logging "True", which means that window considering itself active, which is totally wrong. – BOOtak Mar 21 '15 at 07:12
I don't know python, so I can't help with debugging. – Dmitry Sazonov Mar 21 '15 at 07:35

score 0 · Answer 3 · answered Mar 20 '15 at 13:22

0

You may try isVisible() && isActiveWindow() && !isMinimized()

answered Mar 20 '15 at 13:22

Matt

13,674
1
18
27

isVisible() used to determine is Qwidget visible in terms of calling show()/hide() on it. isMinimized() has no effect, too. – BOOtak Mar 20 '15 at 17:16

Qt 5.4.0 on Windows, isActive() called on QWidget in some cases returns obviously wrong value

3 Answers3