Code:
import subprocess process = subprocess.Popen('echo 5') Error:
Traceback (most recent call last): File "test.py", line 3, in <module> process = subprocess.Popen('echo 5') File "/usr/lib64/python2.6/subprocess.py", line 642, in __init__ errread, errwrite) File "/usr/lib64/python2.6/subprocess.py", line 1238, in _execute_child raise child_exception OSError: [Errno 2] No such file or directory Can someone please advise what is the issue with the above code?
rewrite the code as follows
import subprocess process = subprocess.Popen(['echo', '5']) command should be a list
You have to specify executable and arguments either as a list:
subprocess.Popen(['echo', '5']) or as a string and set shell=True:
subprocess.Popen('echo 5', shell=True) The first is recommended, as it handles needed escaping for you. The second simply passes the string to the shell. According to subprocess docs:
args is required for all calls and should be a string, or a sequence of program arguments. Providing a sequence of arguments is generally preferred, as it allows the module to take care of any required escaping and quoting of arguments (e.g. to permit spaces in file names). If passing a single string, either shell must be True (see below) or else the string must simply name the program to be executed without specifying any arguments.
Popenexpects list["ls", "-lrt"](if you don't useshell=True) - check in documentation.