storing the output of ls command in a shell variable

Question

I am using below command to find a most recent file with name "candump"

ls *.log | grep "candump" | tail -n 1

The output is "candump-2018-04-19_131908.log"

I want to store the output filename to a variable in my shell script. I am using the below commands:

logfile = `ls *.log | grep "candump" | tail -n 1`

and

logfile = $(ls *.log | grep "candump" | tail -n 1)

However, both times I am getting the same error, "logfile: command not found". Am I doing something wrong? Any help is appreciated.

A little search goes a long way. Possible duplicate of How to set a variable to the output from a command in Bash?, How can I assign the output of a command to a shell variable?, Storing output of command in shell variable, etc. — jww
– jww, Commented Apr 22, 2018 at 14:19

Sébastien S. · Accepted Answer · 2018-04-22 03:59:50Z

15

You have to stick the variable name and its value (no space before and after the =).

Try :

logfile=$(ls *.log | grep "candump" | tail -n 1)

answered Apr 22, 2018 at 3:59

Sébastien S.

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1 Comment

Verma Over a year ago

Thanks for the answer, was a silly mistake on my part.

M. Hamza Rajput · Accepted Answer · 2022-09-07 08:57:53Z

0

This is working for me.

#!/bin/bash 

my_command='ls | grep server.js | wc -l';

my_data=$(eval "$my_command");

echo "value in echo is:" $my_data;

if [ $my_data == "1" ]; then
    echo "value is equal to 1";
else
    echo "value is not equal to 1";
fi

answered Sep 7, 2022 at 8:57

M. Hamza Rajput

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