Escape $ dollar sign in variable and execution of command

Asked Mar 16 '20 at 08:03

Active Mar 16 '20 at 08:33

Viewed 4,344 times

I have the following code:

startCmnd="sudo -- bash -c"
test="${startCmnd} \"echo 'Hello this is a \$TEST' >> hello.txt\""
bash -c "${test}"

The expected result must be the text Hello this is a $TEST inside the file, but instead of that, it only put Hello this is a.

It has to be done in that way

Does anyone has any idea of how I fix this or a better way to accomplish this?

Any help it would be preciated

edited Mar 16 '20 at 08:33

asked Mar 16 '20 at 08:03

Ivan Slipkorn

1

Related, if not a dupe: https://unix.stackexchange.com/questions/444946 – Kusalananda Mar 16 '20 at 08:08
Please explain "isn't working". You've told us what you expected, which is good. But what happened instead? – Chris Davies Mar 16 '20 at 08:24
@Kusalananda not sure about that dupe here – Chris Davies Mar 16 '20 at 08:26
1

@roaima In as much as it's about running a command in a variable. Did not mark it as a dupe though as someone may want to also remove the three levels of child shells that the code is using. – Kusalananda Mar 16 '20 at 08:28
@roaima Thanks for the advice, I have updated the question – Ivan Slipkorn Mar 16 '20 at 08:35
2

This seems to be a quoting issue. I'm guessing it would work with \\\$TEST. – Kusalananda Mar 16 '20 at 08:45
@Kusalananda It works, thanks! I think since I have 2 "quotation" levels I have to escape the escaping character that escapes the $ dollar sign – Ivan Slipkorn Mar 16 '20 at 15:04

0 Answers0