将 ls 命令的 output 存储在 shell 变量中

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.

Answer 1

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)

Answer 2

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