How to run a Python script inside another script and close itself immediately
up vote
-2
down vote
favorite
I have a login frame and a main frame that made from pyqt.
I want to run the main command after I clicked the button and when the main frame runs close the login frame.
I wrote a def for clicking button but when the command executed the main frame still waits until I close the main window.
This is the function of clicking the button:
def ButtonClicked(self):
os.system('py Main.py')
self.os.exit()
How can I make it close immediately and not wait for main.py to exit?
python pyqt5
edited Nov 20 at 12:29
usr2564301
17.5k73270
asked Nov 20 at 10:41
ghostDs
85
add a comment |
up vote
-2
down vote
favorite
I have a login frame and a main frame that made from pyqt.
I want to run the main command after I clicked the button and when the main frame runs close the login frame.
I wrote a def for clicking button but when the command executed the main frame still waits until I close the main window.
This is the function of clicking the button:
def ButtonClicked(self):
os.system('py Main.py')
self.os.exit()
How can I make it close immediately and not wait for main.py to exit?
python pyqt5
edited Nov 20 at 12:29
usr2564301
17.5k73270
asked Nov 20 at 10:41
ghostDs
85
start "" py Main.py
– Peter Wood
Nov 20 at 10:47
i dont get it what is the start
– ghostDs
Nov 20 at 10:51
os.system('start "" py Main.py')
– Peter Wood
Nov 20 at 11:12
@PeterWood thanks alot it worked ,but i have a little bug that is when the os.system('start "" py Main.py') runs the console start but not comes popop. i should press it on taskbar to show
– ghostDs
Nov 20 at 11:28
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I have a login frame and a main frame that made from pyqt.
I want to run the main command after I clicked the button and when the main frame runs close the login frame.
I wrote a def for clicking button but when the command executed the main frame still waits until I close the main window.
This is the function of clicking the button:
def ButtonClicked(self):
os.system('py Main.py')
self.os.exit()
How can I make it close immediately and not wait for main.py to exit?
python pyqt5
edited Nov 20 at 12:29
usr2564301
17.5k73270
asked Nov 20 at 10:41
ghostDs
85
I have a login frame and a main frame that made from pyqt.
I want to run the main command after I clicked the button and when the main frame runs close the login frame.
I wrote a def for clicking button but when the command executed the main frame still waits until I close the main window.
This is the function of clicking the button:
def ButtonClicked(self):
os.system('py Main.py')
self.os.exit()
How can I make it close immediately and not wait for main.py to exit?
python pyqt5
python pyqt5
edited Nov 20 at 12:29
usr2564301
17.5k73270
asked Nov 20 at 10:41
ghostDs
85
edited Nov 20 at 12:29
usr2564301
17.5k73270
asked Nov 20 at 10:41
ghostDs
85
edited Nov 20 at 12:29
usr2564301
17.5k73270
edited Nov 20 at 12:29
usr2564301
17.5k73270
edited Nov 20 at 12:29
usr2564301
17.5k73270
17.5k73270
asked Nov 20 at 10:41
ghostDs
85
asked Nov 20 at 10:41
ghostDs
85
asked Nov 20 at 10:41
ghostDs
85
85
start "" py Main.py
– Peter Wood
Nov 20 at 10:47
i dont get it what is the start
– ghostDs
Nov 20 at 10:51
os.system('start "" py Main.py')
– Peter Wood
Nov 20 at 11:12
@PeterWood thanks alot it worked ,but i have a little bug that is when the os.system('start "" py Main.py') runs the console start but not comes popop. i should press it on taskbar to show
– ghostDs
Nov 20 at 11:28
add a comment |
start "" py Main.py
– Peter Wood
Nov 20 at 10:47
i dont get it what is the start
– ghostDs
Nov 20 at 10:51
os.system('start "" py Main.py')
– Peter Wood
Nov 20 at 11:12
@PeterWood thanks alot it worked ,but i have a little bug that is when the os.system('start "" py Main.py') runs the console start but not comes popop. i should press it on taskbar to show
– ghostDs
Nov 20 at 11:28
start "" py Main.py– Peter Wood
Nov 20 at 10:47
start "" py Main.py– Peter Wood
Nov 20 at 10:47
i dont get it what is the start
– ghostDs
Nov 20 at 10:51
i dont get it what is the start
– ghostDs
Nov 20 at 10:51
os.system('start "" py Main.py')– Peter Wood
Nov 20 at 11:12
os.system('start "" py Main.py')– Peter Wood
Nov 20 at 11:12
@PeterWood thanks alot it worked ,but i have a little bug that is when the os.system('start "" py Main.py') runs the console start but not comes popop. i should press it on taskbar to show
– ghostDs
Nov 20 at 11:28
@PeterWood thanks alot it worked ,but i have a little bug that is when the os.system('start "" py Main.py') runs the console start but not comes popop. i should press it on taskbar to show
– ghostDs
Nov 20 at 11:28
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Not sure if your approach is the best here, but just to answer your specific question: os.system is not the right function here since it waits for the output of the call before continuing (as you might have noticed...). Try to "spawn" a process and tell Python not to wait for any result with something like this:
os.spawnl(os.P_DETACH, 'py Main.py')
EDIT
Given the comments I'll change the answer a bit. To use spawnl in a "with path" mode you have to add a p at the end (spawnlp()). Although the os.system and os.spawn** provide basic functionality, the docs refer to the subprocess library for better control. It makes stuff a little more complicated but I think this example does what you ask for:
import sys
import subprocess
# Constant for detaching a process
DETACHED_PROCESS = 0x00000008
# Start the process and get its PID
pid = subprocess.Popen(["python", "you_second_script.py"], close_fds=True, creationflags=DETACHED_PROCESS).pid
print(pid)
# The end of this script, the started process will continue
sys.exit()
answered Nov 20 at 12:25
Gijs Wobben
365
os.spawnldoesn't usePATHto locatepy
– Peter Wood
Nov 20 at 19:50
thanks for the answer , but @PeterWood answer helped me through this
– ghostDs
Nov 22 at 8:24
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Not sure if your approach is the best here, but just to answer your specific question: os.system is not the right function here since it waits for the output of the call before continuing (as you might have noticed...). Try to "spawn" a process and tell Python not to wait for any result with something like this:
os.spawnl(os.P_DETACH, 'py Main.py')
EDIT
Given the comments I'll change the answer a bit. To use spawnl in a "with path" mode you have to add a p at the end (spawnlp()). Although the os.system and os.spawn** provide basic functionality, the docs refer to the subprocess library for better control. It makes stuff a little more complicated but I think this example does what you ask for:
import sys
import subprocess
# Constant for detaching a process
DETACHED_PROCESS = 0x00000008
# Start the process and get its PID
pid = subprocess.Popen(["python", "you_second_script.py"], close_fds=True, creationflags=DETACHED_PROCESS).pid
print(pid)
# The end of this script, the started process will continue
sys.exit()
answered Nov 20 at 12:25
Gijs Wobben
365
os.spawnldoesn't usePATHto locatepy
– Peter Wood
Nov 20 at 19:50
thanks for the answer , but @PeterWood answer helped me through this
– ghostDs
Nov 22 at 8:24
add a comment |
up vote
1
down vote
Not sure if your approach is the best here, but just to answer your specific question: os.system is not the right function here since it waits for the output of the call before continuing (as you might have noticed...). Try to "spawn" a process and tell Python not to wait for any result with something like this:
os.spawnl(os.P_DETACH, 'py Main.py')
EDIT
Given the comments I'll change the answer a bit. To use spawnl in a "with path" mode you have to add a p at the end (spawnlp()). Although the os.system and os.spawn** provide basic functionality, the docs refer to the subprocess library for better control. It makes stuff a little more complicated but I think this example does what you ask for:
import sys
import subprocess
# Constant for detaching a process
DETACHED_PROCESS = 0x00000008
# Start the process and get its PID
pid = subprocess.Popen(["python", "you_second_script.py"], close_fds=True, creationflags=DETACHED_PROCESS).pid
print(pid)
# The end of this script, the started process will continue
sys.exit()
answered Nov 20 at 12:25
Gijs Wobben
365
os.spawnldoesn't usePATHto locatepy
– Peter Wood
Nov 20 at 19:50
thanks for the answer , but @PeterWood answer helped me through this
– ghostDs
Nov 22 at 8:24
add a comment |
up vote
1
down vote
up vote
1
down vote
Not sure if your approach is the best here, but just to answer your specific question: os.system is not the right function here since it waits for the output of the call before continuing (as you might have noticed...). Try to "spawn" a process and tell Python not to wait for any result with something like this:
os.spawnl(os.P_DETACH, 'py Main.py')
EDIT
Given the comments I'll change the answer a bit. To use spawnl in a "with path" mode you have to add a p at the end (spawnlp()). Although the os.system and os.spawn** provide basic functionality, the docs refer to the subprocess library for better control. It makes stuff a little more complicated but I think this example does what you ask for:
import sys
import subprocess
# Constant for detaching a process
DETACHED_PROCESS = 0x00000008
# Start the process and get its PID
pid = subprocess.Popen(["python", "you_second_script.py"], close_fds=True, creationflags=DETACHED_PROCESS).pid
print(pid)
# The end of this script, the started process will continue
sys.exit()
answered Nov 20 at 12:25
Gijs Wobben
365
Not sure if your approach is the best here, but just to answer your specific question: os.system is not the right function here since it waits for the output of the call before continuing (as you might have noticed...). Try to "spawn" a process and tell Python not to wait for any result with something like this:
os.spawnl(os.P_DETACH, 'py Main.py')
EDIT
Given the comments I'll change the answer a bit. To use spawnl in a "with path" mode you have to add a p at the end (spawnlp()). Although the os.system and os.spawn** provide basic functionality, the docs refer to the subprocess library for better control. It makes stuff a little more complicated but I think this example does what you ask for:
import sys
import subprocess
# Constant for detaching a process
DETACHED_PROCESS = 0x00000008
# Start the process and get its PID
pid = subprocess.Popen(["python", "you_second_script.py"], close_fds=True, creationflags=DETACHED_PROCESS).pid
print(pid)
# The end of this script, the started process will continue
sys.exit()
answered Nov 20 at 12:25
Gijs Wobben
365
edited Nov 22 at 7:20
answered Nov 20 at 12:25
Gijs Wobben
365
answered Nov 20 at 12:25
Gijs Wobben
365
answered Nov 20 at 12:25
Gijs Wobben
365
365
os.spawnldoesn't usePATHto locatepy
– Peter Wood
Nov 20 at 19:50
thanks for the answer , but @PeterWood answer helped me through this
– ghostDs
Nov 22 at 8:24
add a comment |
os.spawnldoesn't usePATHto locatepy
– Peter Wood
Nov 20 at 19:50
thanks for the answer , but @PeterWood answer helped me through this
– ghostDs
Nov 22 at 8:24
os.spawnl doesn't use PATH to locate py– Peter Wood
Nov 20 at 19:50
os.spawnl doesn't use PATH to locate py– Peter Wood
Nov 20 at 19:50
thanks for the answer , but @PeterWood answer helped me through this
– ghostDs
Nov 22 at 8:24
thanks for the answer , but @PeterWood answer helped me through this
– ghostDs
Nov 22 at 8:24
add a comment |
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start "" py Main.py– Peter Wood
Nov 20 at 10:47
i dont get it what is the start
– ghostDs
Nov 20 at 10:51
os.system('start "" py Main.py')– Peter Wood
Nov 20 at 11:12
@PeterWood thanks alot it worked ,but i have a little bug that is when the os.system('start "" py Main.py') runs the console start but not comes popop. i should press it on taskbar to show
– ghostDs
Nov 20 at 11:28