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[PyGUI] Failed when communicating between threads
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[PyGUI] Failed when communicating between threads
#2
I don't think signals works between two threads, they only work between main GUI and other threads.

on_GUI_changeValue function is never called as thread.start() will only execute run method.

To communicate between threads you can use global variables or queues.

(Sep-11-2017, 11:10 AM)pythonisse Wrote: "QPixmap: It is not safe to use pixmaps outside the GUI thread"

What they mean by GUI thread is the main thread that is started with .show() method.
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RE: Failed when communicating between threads - by hbknjr - Sep-11-2017, 12:43 PM

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