I am working on a program that requires to call another python script and truncate the execution of the current file. I tried doing the same using the os.close() function. As follows:
def call_otherfile(self):
os.system("python file2.py") #Execute new script
os.close() #close Current Script
Using the above code I am able to open the second file but am unable to close the current one.I know I am silly mistake but unable to figure out what's it.
To do this you will need to spawn a subprocess directly. This can either be done with a more low-level fork and exec model, as is traditional in Unix, or with a higher-level API like subprocess
.
import subprocess
import sys
def spawn_program_and_die(program, exit_code=0):
"""
Start an external program and exit the script
with the specified return code.
Takes the parameter program, which is a list
that corresponds to the argv of your command.
"""
# Start the external program
subprocess.Popen(program)
# We have started the program, and can suspend this interpreter
sys.exit(exit_code)
spawn_program_and_die(['python', 'path/to/my/script.py'])
# Or, as in OP's example
spawn_program_and_die(['python', 'file2.py'])
Also, just a note on your original code. os.close
corresponds to the Unix syscall close
, which tells the kernel that your program that you no longer need a file descriptor. It is not supposed to be used to exit the program.
If you don't want to define your own function, you could always just call subprocess.Popen
directly like Popen(['python', 'file2.py'])
Traceback (most recent call last): File "v8function_handler.pyx", line 48, in cefpython_py27.V8FunctionHandler_Execute File "file1.py", line 62, in call_otherfile subprocess.Popen("python file2.py") File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 710, in init errread, errwrite) File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 1335, in _execute_child raise child_exception OSError: [Errno 2] No such file or directory
got the above error on using subprocess.Popen() however same works fine with os.system().
Try it with
['python', 'file2.py']
.os.system
takes the entire command you would run in shell.subprocess
takes the full argv as a list.Popen('python file2.py')
is looking for a program on your path namedpython file2.py