command line working but not working in shell script

Question

I use the following command to get the output (number tagged in the latest file):

 ls -lrt | tail -n1 |  awk -F'.' '{print $4}' | grep -o '[0-9]\+'

but when I use the same command in a shell script and get the result in a variable c, I don't get the proper result.

My script:

#! /bin/sh

c=ls -lrt | tail -n1 |  awk -F'.' '{print $4}' | grep -o '[0-9]\+'
echo $c

The error that shows when I execute is:

 -lrt: command not found

Does this answer your question? How do I set a variable to the output of a command in Bash? — Benjamin W.
– Benjamin W., Commented May 26, 2020 at 12:38

Benjamin W. · Accepted Answer · 2020-05-26 12:40:08Z

1

What you are doing: trying to run a command with a per-command variable.

VAR=foo ./somecommand.sh

would run ./somecommand.sh with VAR set to foo. But VAR would not be set for commands executed later.

Your example sets c=ls and tries to run -lrt - obviously this is neither found nor intended.

I assume you want to save the output to variable c.

You would use

c=`ls -lrt | tail -n1 | awk -F'.' '{print $4}' | grep -o '[0-9]+'`
echo "$c"

Modern shells use $(...) instead of `...` for command substitution, as nesting command substitution gets easier with this syntax.

Benjamin W.

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answered May 26, 2020 at 11:34

user3188140

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