Echo variable using sudo bash -c 'echo $myVariable' - bash script

Question

I want to echo a string into the /etc/hosts file. The string is stored in a variable called $myString.

When I run the following code the echo is empty:

finalString="Hello\nWorld"
sudo bash -c 'echo -e "$finalString"'

What am I doing wrong?

Ignacio Vazquez-Abrams · Accepted Answer · 2016-05-03T12:48:30.733

6

You're not exporting the variable into the environment so that it can be picked up by subprocesses.
You haven't told sudo to preserve the environment.

\

finalString="Hello\nWorld"
export finalString
sudo -E bash -c 'echo -e "$finalString"'

Alternatively, you can have the current shell substitute instead:

finalString="Hello\nWorld"
sudo bash -c 'echo -e "'"$finalString"'"'

edited May 03 '16 at 12:48

answered May 03 '16 at 12:39

That is: using `export` makes variables available in the sub-shells. – fedorqui May 03 '16 at 12:41
to export the variable into the environment is it just 'export finalString'? Because it's still not printing the variable. – user2771150 May 03 '16 at 12:43
@user2771150 you probably need `-E` in sudo --> `sudo -E bash -c 'echo -e "$finalString"'`. – fedorqui May 03 '16 at 12:48
are they both 'safe'? is there a 'recommended' way? sorry. I'm pretty new to bash scripting. – user2771150 May 03 '16 at 12:51
if you echo to a file with sudo bash -c 'echo -e "'"$finalString"'" > your_text_file', then it is better to go for the alternative – HanniBaL90 Oct 05 '22 at 09:41

Jahid · Answer 2 · 2016-05-03T12:53:07.150

1

You can do this:

bash -c "echo -e '$finalString'"

i.e using double quote to pass argument to the subshell, thus the variable ($finalString) is expanded (by the current shell) as expected.

Though I would recommend not using the -e flag with echo. Instead you can just do:

finalString="Hello
World"

edited May 03 '16 at 12:53

answered May 03 '16 at 12:47

Jahid

2 Answers2