I have a file named test's file.txt inside test's dir directory. So the file-path becomes test's dir/test's file.txt. I want to cat the content of the file but since the file contains an apostrophe and a space it is giving me hard time achieving that. I have tried several commands including

1. sh -c "cat 'test's dir/test's file.txt'"
2. sh -c 'cat "test's dir/test's file.txt"'
3. sh -c "cat '"'"'test's dir/test's file.txt'"'"'"
4. sh -c 'cat "test\'s\ dir/test\'s\ file.txt"' and many more ... But none of them is working.

Some help would be really appreciated. Thank you!

CodePudding user response：

Would you please try:

sh -c "cat 'test'\''s dir/test'\''s file.txt'"

As for the pathname part, it is a concatenation of:

'test'
\'
's dir/test'
\'
's file.txt'

[Edit]
If you want to execute the shell command in python, would you please try:

#!/usr/bin/python3

import subprocess

path="test's dir/test's file.txt"
subprocess.run(['cat', path])

or immediately:

subprocess.run(['cat', "test's dir/test's file.txt"])

As the subprocess.run() function takes the command as a list, not a single string (possible with shell=True option), we do not have to worry about the extra quoting around the command.

Please note subprocess.run() is supported by Python 3.5 or newer.

CodePudding user response：

You can use here-doc:

sh -s <<-'EOF'
cat "test's dir/test's file.txt"
EOF

CodePudding user response：

This option avoids the need for two levels of quoting:

sh -c 'cat -- "$0"' "test's dir/test's file.txt"

See How to use positional parameters with "bash -c" command?. (It applies to sh -c too.)