“no such file or directory” in python script calling other python script in debian linux

Question

Hi I've got the following written in a python script pythonscript1.py in linux debian in the directory /home/user/a/:

import subprocess
subprocess.call(["python" /home/user/b/phpcall.py"])

where the phpcall.py script contains:

import subprocess
subprocess.call(["php", "/home/user/b/phpscript1.php"])

individually called from console all scripts function perfectly, but when I use the first script whilst the 2nd script calls/looks for a file in directory b, rather than a, It yields the following error:

"PHP warning: include_once(something.php): failed to open stream: no such file in /home/user/b/phpschript1.php on line 25

Now it is quite clear to me the problem is that it cannot reach out of it's original directory. But I don't know what command I should add to the first script to allow the 2nd script to look in folder b.

So far google results have suggested something with "include_path='include'" but I don't know how/where to incorporate the statement succesfully.

Any suggestions on the correct syntax would be appreciated!

Answer 1

For future reference the code that directly called the php script from python within was:

import os
os.chdir("/home/user/b")
os.system("php /home/user/b/phpscript1.php")

.Thanks Marc B & I love the dynamic approach Sebastian, thank you :)

Answer 2

you can try this program:

import subprocess
import sys
proc = subprocess.Popen(
       [sys.executable, '/home/user/b/phpcall.py'],stdin=subprocess.PIPE)

proc.communicate()

i think it's work. if not let know.

Answer 3

If the php script works only if you start it from the directory with the python script then you could use cwd parameter to change the working directory for the subprocess:

#!/usr/bin/env python
from subprocess import check_call

check_call(["php", "/home/user/b/phpscript1.php"], cwd="/home/user/b")

I've added the shebang #!/usr/bin/env python so that you could run the script directly as /path/to/phpcall.py assuming it is executable ( chmod +x phpcall.py ).

Note: if you want to run the python script as user user then you could specify the path using ~ :

#!/usr/bin/env python
import os
from subprocess import check_call

script_dir = os.path.expanduser("~/b")
check_call(["php", os.path.join(script_dir, "phpscript1.php")], cwd=script_dir)

To avoid hardcoding the script directory, you could find it dynamically:

#!/usr/bin/env python
import os
from subprocess import check_call

script_dir = get_script_dir()
check_call(["php", os.path.join(script_dir, "phpscript1.php")], cwd=script_dir)

where get_script_dir() .