Execute a file line change

Question

I currently have an automated process that allows me to run a python script to be able to edit the .bashrc file in my vagrant box.

The code looks like this:

if 'THIS STRING' not in open('/home/vagrant/.bashrc').read():
    with open("/home/vagrant/.bashrc", "a") as bash:
        bash.write('THIS STRING')
    execfile("/home/vagrant/.bashrc")

The code THIS STRING exists in the .bashrc and if it doesnt, it should open it, add it in there and source the file.

To source the file, i thought i would use the execfile function in python. This can definitely be done just by sourcing it manually;

$ sudo source /home/vagrant/.bashrc

However, i am getting an error message stating the following;

Traceback (most recent call last):
  File "/var/www/soething.py", line 233, in <module>
    execfile("/home/vagrant/.bashrc")
  File "<string>", line 6
    case $- in

The error message is targeting the execfile every single time.

I have also tried the following;

exec(open("/home/vagrant/.bashrc").read())

Still the same error - Any thoughts on why this error or what i am doing wrong?

`exec` and `execfile` are used to execute python files, not bash scripts. — Goodies, Jan 02 '18 at 18:42
@Goodies so i wouldnt be able to source a file after making edits to it through python? — Sophie Rhodes, Jan 02 '18 at 18:44
I think this is a dup of https://stackoverflow.com/questions/3503719/emulating-bash-source-in-python — sehafoc, Jan 02 '18 at 18:47
You can't source a file within Python: https://stackoverflow.com/questions/3661566/python-source-home-bashrc-with-os-system — Nelson Yeung, Jan 02 '18 at 18:53

Execute a file line change

0 Answers0