Comments (8)
I'm sorry but this is not enough information to provide any help. Is it possible to create example code, that is as small as possible, but that I can run and that demonstrates the problem?
from flexx.
ok, this is a complete example, you need
You need to package an application through the freeze method, then run, click this button.
import time
from multiprocessing import Pool
from flexx import flx
def down_single(index):
print(f"{index}")
time.sleep(index)
def down_multi():
executor = Pool(2)
for i in range(4):
executor.apply_async(down_single, args=(i, ))
executor.close()
executor.join()
class Example(flx.PyWidget):
def init(self):
with flx.VBox():
self.lable = flx.MultiLineEdit()
self.button = flx.Button(text="test")
flx.HBox(flex=1)
@flx.reaction("button.pointer_click")
def down_btn(self, *events):
self.down_ac()
@flx.action
def down_ac(self):
down_multi()
self.lable.set_text("success!")
from flexx.
After running, multiple windows will be launched and cannot be closed, as if in a computer virus, hahaha
from flexx.
Did you put the code to launch the app in an if __name__ == "__main__"
?
if __name__ == '__main__':
m = flx.App(Example).launch('firefox')
flx.run()
from flexx.
Here's the code I packed
def test_make_package_3(self):
flx.App(Example).freeze("./", excludes=[], launch="app")
from flexx.
Oh, I have no idea how multiprocessing works together with frozen apps. But I do know that when you "launch the pool", it spawns new processes using the same __main__
module, so if you don't guard with if __name__ == "__main__"
it will re-launch the app again, and again ...
from flexx.
yeah,i run with if name == "main", the program running ok,hope this problem can be solved later. thank you
from flexx.
hello,
Pyinstaller packages programs that use multiprocessing must call the multiprocessing.freeze_support () method to freeze the process. I try to write the generated examp.py in the following format to solve this problem.
pyinstaller.exe "E:\code\flexx\learning\example.spec"
from flexx.
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from flexx.