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How to restart a Python script?

In a program I am writing in python I need to completely restart the program if a variable becomes true, looking for a while I found this command:
while True:
if reboot == True:
os.execv(sys.argv[0], sys.argv)
When executed it returns the error [Errno 8] Exec format error. I searched for further documentation on os.execv, but didn't find anything relevant, so my question is if anyone knows what I did wrong or knows a better way to restart a script (by restarting I mean completely re-running the script, as if it were been opened for the first time, so with all unassigned variables and no thread running).

There are multiple ways to achieve the same thing. Start by modifying the program to exit whenever the flag turns True. Then there are various options, each one with its advantages and disadvantages.
Wrap it using a bash script.
The script should handle exits and restart your program. A really basic version could be:
#!/bin/bash
while :
do
python program.py
sleep 1
done
Start the program as a sub-process of another program.
Start by wrapping your program's code to a function. Then your __main__ could look like this:
def program():
### Here is the code of your program
...
while True:
from multiprocessing import Process
process = Process(target=program)
process.start()
process.join()
print("Restarting...")
This code is relatively basic, and it requires error handling to be implemented.
Use a process manager
There are a lot of tools available that can monitor the process, run multiple processes in parallel and automatically restart stopped processes. It's worth having a look at PM2 or similar.
IMHO the third option (process manager) looks like the safest approach. The other approaches will have edge cases and require implementation from your side to handle edge cases.

This has worked for me. Please add the shebang at the top of your code and os.execv() as shown below
#!/usr/bin/env python3
import os
import sys
if __name__ == '__main__':
while True:
reboot = input('Enter:')
if reboot == '1':
sys.stdout.flush()
os.execv(sys.executable, [sys.executable, __file__] + [sys.argv[0]])
else:
print('OLD')

I got the same "Exec Format Error", and I believe it is basically the same error you get when you simply type a python script name at the command prompt and expect it to execute. On linux it won't work because a path is required, and the execv method is basically encountering the same error.
You could add the pathname of your python compiler, and that error goes away, except that the name of your script then becomes a parameter and must be added to the argv list. To avoid that, make your script independently executable by adding "#!/usr/bin/python3" to the top of the script AND chmod 755.
This works for me:
#!/usr/bin/python3
# this script is called foo.py
import os
import sys
import time
if (len(sys.argv) >= 2):
Arg1 = int(sys.argv[1])
else:
sys.argv.append(None)
Arg1 = 1
print(f"Arg1: {Arg1}")
sys.argv[1] = str(Arg1 + 1)
time.sleep(3)
os.execv("./foo.py", sys.argv)
Output:
Arg1: 1
Arg1: 2
Arg1: 3
.
.
.

Python subprocess shell scripts still runs in background

I am running two python scripts using subprocess one of them still runs.
import subprocess
subprocess.run("python3 script_with_loop.py & python3 scrip_with_io.py", shell=True)
script_with_loop still runs in the background.
What is the way to kill both scripts if one of them dies?

So, you're basically not using python here, you're using your shell.
a & b runs a, disavows it, and runs b. Since you're using the shell, if you wanted to terminate the background task, you'd have to use shell commands to do that.
Of course, since you're using python, there is a better way.
with subprocess.Popen(["somecommand"]) as proc:
try:
subprocess.run(["othercommand"])
finally:
proc.terminate()
Looking at your code though - python3 script_with_loop.py and python3 script_with_io.py - my guess is you'd be better off using the asyncio module because it basically does what the names of those two files are describing.

you should use threading for this sort of thing. try this.
import threading
def script_with_loop():
try:
# script_with_loop.py code goes here
except:
_exit()
def script_with_io():
try:
# script_with_io.py code goes here
except:
_exit()
threading.Thread(target=script_with_loop, daemon=True).start()
threading.Thread(target=script_with_io, daemon=True).start()

Python subprocess always waits for programm [duplicate]

I'm trying to port a shell script to the much more readable python version. The original shell script starts several processes (utilities, monitors, etc.) in the background with "&". How can I achieve the same effect in python? I'd like these processes not to die when the python scripts complete. I am sure it's related to the concept of a daemon somehow, but I couldn't find how to do this easily.

While jkp's solution works, the newer way of doing things (and the way the documentation recommends) is to use the subprocess module. For simple commands its equivalent, but it offers more options if you want to do something complicated.
Example for your case:
import subprocess
subprocess.Popen(["rm","-r","some.file"])
This will run rm -r some.file in the background. Note that calling .communicate() on the object returned from Popen will block until it completes, so don't do that if you want it to run in the background:
import subprocess
ls_output=subprocess.Popen(["sleep", "30"])
ls_output.communicate() # Will block for 30 seconds
See the documentation here.
Also, a point of clarification: "Background" as you use it here is purely a shell concept; technically, what you mean is that you want to spawn a process without blocking while you wait for it to complete. However, I've used "background" here to refer to shell-background-like behavior.

Note: This answer is less current than it was when posted in 2009. Using the subprocess module shown in other answers is now recommended in the docs
(Note that the subprocess module provides more powerful facilities for spawning new processes and retrieving their results; using that module is preferable to using these functions.)
If you want your process to start in the background you can either use system() and call it in the same way your shell script did, or you can spawn it:
import os
os.spawnl(os.P_DETACH, 'some_long_running_command')
(or, alternatively, you may try the less portable os.P_NOWAIT flag).
See the documentation here.

You probably want the answer to "How to call an external command in Python".
The simplest approach is to use the os.system function, e.g.:
import os
os.system("some_command &")
Basically, whatever you pass to the system function will be executed the same as if you'd passed it to the shell in a script.

I found this here:
On windows (win xp), the parent process will not finish until the longtask.py has finished its work. It is not what you want in CGI-script. The problem is not specific to Python, in PHP community the problems are the same.
The solution is to pass DETACHED_PROCESS Process Creation Flag to the underlying CreateProcess function in win API. If you happen to have installed pywin32 you can import the flag from the win32process module, otherwise you should define it yourself:
DETACHED_PROCESS = 0x00000008
pid = subprocess.Popen([sys.executable, "longtask.py"],
creationflags=DETACHED_PROCESS).pid

Use subprocess.Popen() with the close_fds=True parameter, which will allow the spawned subprocess to be detached from the Python process itself and continue running even after Python exits.
https://gist.github.com/yinjimmy/d6ad0742d03d54518e9f
import os, time, sys, subprocess
if len(sys.argv) == 2:
time.sleep(5)
print 'track end'
if sys.platform == 'darwin':
subprocess.Popen(['say', 'hello'])
else:
print 'main begin'
subprocess.Popen(['python', os.path.realpath(__file__), '0'], close_fds=True)
print 'main end'

Both capture output and run on background with threading
As mentioned on this answer, if you capture the output with stdout= and then try to read(), then the process blocks.
However, there are cases where you need this. For example, I wanted to launch two processes that talk over a port between them, and save their stdout to a log file and stdout.
The threading module allows us to do that.
First, have a look at how to do the output redirection part alone in this question: Python Popen: Write to stdout AND log file simultaneously
Then:
main.py
#!/usr/bin/env python3
import os
import subprocess
import sys
import threading
def output_reader(proc, file):
while True:
byte = proc.stdout.read(1)
if byte:
sys.stdout.buffer.write(byte)
sys.stdout.flush()
file.buffer.write(byte)
else:
break
with subprocess.Popen(['./sleep.py', '0'], stdout=subprocess.PIPE, stderr=subprocess.PIPE) as proc1, \
subprocess.Popen(['./sleep.py', '10'], stdout=subprocess.PIPE, stderr=subprocess.PIPE) as proc2, \
open('log1.log', 'w') as file1, \
open('log2.log', 'w') as file2:
t1 = threading.Thread(target=output_reader, args=(proc1, file1))
t2 = threading.Thread(target=output_reader, args=(proc2, file2))
t1.start()
t2.start()
t1.join()
t2.join()
sleep.py
#!/usr/bin/env python3
import sys
import time
for i in range(4):
print(i + int(sys.argv[1]))
sys.stdout.flush()
time.sleep(0.5)
After running:
./main.py
stdout get updated every 0.5 seconds for every two lines to contain:
0
10
1
11
2
12
3
13
and each log file contains the respective log for a given process.
Inspired by: https://eli.thegreenplace.net/2017/interacting-with-a-long-running-child-process-in-python/
Tested on Ubuntu 18.04, Python 3.6.7.

You probably want to start investigating the os module for forking different threads (by opening an interactive session and issuing help(os)). The relevant functions are fork and any of the exec ones. To give you an idea on how to start, put something like this in a function that performs the fork (the function needs to take a list or tuple 'args' as an argument that contains the program's name and its parameters; you may also want to define stdin, out and err for the new thread):
try:
pid = os.fork()
except OSError, e:
## some debug output
sys.exit(1)
if pid == 0:
## eventually use os.putenv(..) to set environment variables
## os.execv strips of args[0] for the arguments
os.execv(args[0], args)

You can use
import os
pid = os.fork()
if pid == 0:
Continue to other code ...
This will make the python process run in background.

I haven't tried this yet but using .pyw files instead of .py files should help. pyw files dosen't have a console so in theory it should not appear and work like a background process.

Execute and wait for results C program and Python script

My today's task is to create a Python script (say A.py) which can do the following things:
Start a C program (say CProg) passing some params
Start another Python script (say B.py) passing other params
Join/Wait until B.py has finished
Send a SIGINT to CProg
Iterate (this won't be a problem at all I think :P)
Since I'm pretty new in developing Python scripts and my mind is quite full of C/C++ thread/join/execve/... I'd like to ask you if there's a proper way to accomplish my task. I've read some related topics on SO (some talk about PIPEs or Execl) but I'm not sure what to use yet.
Thanks in advance

Use subprocess module.
import os
import signal
import subprocess
import sys
params = [...]
for param for params:
proc = subprocess.Popen(['/path/to/CProg', param.., param..])
subprocess.call([sys.executable, 'B.py', param.., param...])
os.kill(proc.pid, signal.SIGINT)
proc.wait()

why process doesn't join and doesn't run?

i have a simple problem to solve(more or less)
if i watch python multiprocessing tutorials i see that a process should be started more or less like this:
from multiprocessing import *
def u(m):
print(m)
return
A=Process(target=u,args=(0,))
A.start()
A.join()
It should print a 0 but nothing gets printed. Instead it hangs forever at the A.join().
if i manually start the function u doing this
A.run()
it actually prints 0 on the shell but it doesn't work simultaneously
for example the output of following code:
from multiprocessing import *
from time import sleep
def u(m):
sleep(1)
print(m)
return
A=Process(target=u,args=(1,))
A.start()
print(0)
should be
0
1
but actually is
0
and if i add before the last line
A.run()
then the output becomes
1
0
this seems confusing to me...
and if i try to join the process it waits forever.
however,if it can help giving me an answer
my OS is Mac os x 10.6.8
python versions used are 3.1 and 3.3
my computer has 1 intel core i3 processor
--Update--
I have noticed that this strange behaviour is present only when launching the program from IDLE ,if i run the program from the terminal everything works as it is supposed to,so this problem must be connected to some IDLE bug.
But runnung programs from terminal is even weirder: using something like range(100000000) activates all my computer's ram until the end of the program; if i remember well this shouldn't happen in python 3,only in older python versions.
I hope these new informations will help you giving an answer
--Update 2--
the bug occurs even if i don't perform output from my process,because setting this:
def u():
return
as the target of the process and then starting it , if i try to join the process,idle waits forever

As suggested here and here, the problem is that IDLE overrides sys.stdin and sys.stdout in some weird ways, which do not propagate cleanly to processes you spawn from it (they are not real filehandles).
The first link also indicates it's unlikely to be fixed any time soon ("may be a 'cannot fix' issue", they say).
So unfortunately the only solution I can suggest is not to use IDLE for this script...

Have you tried adding A.join() to your program? I am guessing that your main process is exiting before the child process prints which is causing the output to be hidden. If you tell the main process to wait for the child process (A.join()), I bet you'll see the output you expect.

Given that it only happens with IDLE, I suspect the problem has to do with the stdout used by both processes. Perhaps it's some file-like object that's not safe to use from two different processes.
If you don't have the child process write to stdout, I suspect it will complete and join properly. For example, you could have it write to a file, instead. Or you could set up a pipe between the parent and child.

Have you tried unbuffered output? Try importing the sys module and change the print statement:
print >> sys.stderr, m
How does this affect the behavior? I'm with the others that suspect that IDLE is mucking with the stdio . . .