使用Python 3 Idle运行.sh文件

Question

I am trying run a sh file with python 3. My .sh file will work on Terminal. My operating system is Raspbian. I try that code :

import time
import os
import subprocess

# STEP 1
text='sudo somecode'
savefile=open('step1.sh','w')
savefile.write(text)
savefile.close()
time.sleep(2)

shellscript=subprocess.Popen(['./step1.sh'], stdin=subprocess.PIPE)

but it not works...

Answer 1

The is undoubtedly a permissions issue. In order to be able to "directly" execute a file (a la "dot slash" - ./yourfile ), the file in question needs the "execute bit" set. Try using ls -l to see the file you've just created with your script. I'll bet the file does not have the execute bit:

$ ls -l ./step.sh
-rw-r--r-- 1 furkan furkan 0 Nov 13 20:51 step.sh

Note the lack of x in that first column. You might chmod to add the execute bit:

$ chmod +x ./step.sh
$ ls -l ./step.sh
-rwxr-xr-x 1 furkan furkan 0 Nov 13 20:51 step.sh

With the execute bit set, you could then use a "dot slash" construct.

However, I doubt you want to execute chmod from within your script, so instead tell your script the actual program you want to run step.sh , namely, sh :

shellscript=subprocess.Popen(['sh', './step1.sh'], stdin=subprocess.PIPE)

Or in the simple case of your example, simply go directly to sudo:

shellscript=subprocess.Popen(['sudo', 'yourexecutable'], stdin=subprocess.PIPE)

Note that if you're being robust, I might consider adding some absolute paths, or ensuring your PATH variable is set. However the key to your problem is the misunderstanding of what 'executable' means.