passing arguments from a python program to other while executing itwith exec() or spawn() in LINUX

[gaurav kashyap]

HI all,
i have two python programs as 1.py and 2.py

1.py
import os
import sys
processID=os.spawnl(os.P_WAIT,'/usr/local/bin/python','python','/
mywork/2.py ' + 'hi')

2.py
import sys
domain= str(sys.argv[1] )
print domain

IN LINUX
while executing 1.py,the argument 'hi' is not passed to the 2.py and
error message is displayed as :
python: can't open file '/mywork/2.py'.If i execute the program from
shell like:
python 2.py hi,then it works fine


IN WINDOWS
this is working fine

PLEASE HELP.
thanks
gaurav

 

[Peter Otten]

HI all,
i have two python programs as 1.py and 2.py

1.py
import os
import sys
processID=os.spawnl(os.P_WAIT,'/usr/local/bin/python','python','/
mywork/2.py ' + 'hi')

2.py
import sys
domain= str(sys.argv[1] )
print domain

IN LINUX
while executing 1.py,the argument 'hi' is not passed to the 2.py and
error message is displayed as :
python: can't open file '/mywork/2.py'.

Did you cut and paste that? I would expect the message to be

python: can't open file '/mywork/2.py hi'.

or similar, i. e. the script name is assumed to be '/mywork/2.py hi'. You
have to pass arguments to the 2.py script as separate arguments to
os.spawnl()

os.spawnl(os.P_WAIT,'/usr/local/bin/python','python','/mywork/2.py', 'hi')

Peter

 

[gaurav kashyap]

Thanks for the help Peter.Its working fine now