Using Variable in Shell script as input for another command

Question

#!/bin/bash

sudo docker-compose -f /home/administrator/compose/docker-compose.yml up --build -d
OUTPUT=$(docker ps | grep 'nginx_custom' | awk '{ print $1 }')

echo $OUTPUT

sudo docker $OUTPUT nginx -s reload

This the the ID that get´s printed correctly in the console.

6e3b3aa3fbc4

This command works fine.

 docker exec 6e3b3aa3fbc4 nginx -s reload

However the variable seems not to get passed to the command here:

sudo docker $OUTPUT nginx -s reload

I am quite unfamiliar with the shell :(. How do I pass the variable to a command that is longer than just echo?

How do you know it doesn't get passed to the command? What error do you see? Have you tried running set -x before issuing sudo docker ... command to see what's going on? — oguz ismail
– oguz ismail, Commented Jun 5, 2020 at 13:30

vgersh99 · Accepted Answer · 2020-06-05 13:34:19Z

2

add set -x to the script and see what happens: you can probably get rid of grep and incorporate it inside awk

#!/bin/bash
set -x
sudo docker-compose -f /home/administrator/compose/docker-compose.yml up --build -d
OUTPUT=$(docker ps | awk '/ngnix_custom/{ print $1 }')

echo $OUTPUT

sudo docker $OUTPUT nginx -s reload

answered Jun 5, 2020 at 13:34

vgersh99

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