Executing bash shell script accepting argument and running in background

I have written a shell script run.sh to trigger a few tasks based on the user’s input

#!/bin/bash

echo "Please choose mode [1-3]: "
read MODE

case $MODE in

  1)
    echo -n "Enter iteration: "
    read TIME
    echo "Start Updating ..."
    task 1 && task 2 && task 3  
    ;;

  2)
    echo -n "Enter Seed Value: "
    read SEED
    echo "Start Updating ..."
    task 4
    task 5
    ;;

  3)
    echo -n "Enter regression minimum value: "
    read MIN
    echo "Start Updating ..."
    task 6
    ;;

  *)
    echo -n "Unknown option - Exit"
    ;;
esac

The tasks 1,2 … 6 are php scripts that are run like /usr/bin/php task1.php $TIME with $TIME as an argument for the php script etc…

The script runs fine when I type bash run.sh but since tasks 1-6 takes a long time to complete I would like an option to run the script in background while I disconnect from the terminal. However if I run the script using bash run.sh & I encountered an error like this:

Please choose mode [1-3]: 2
-bash: 2: command not found

[5]+  Stopped                 bash run.sh

It seems like bash interpreted my input 2 as an argument not corresponding to read MODE but instead of bash run.sh 2 which causes an error. I cannot change the script such that tasks 1-6 are run in the background like task 1 & task 2 & etc. because task 2 can only start running after task 1 is completed.

How can I accomplish what I want to do?

Answer

You could run all of those tasks sequentially in a background subshell

( task 1; task 2; task 3 ) &

Try this out with:

( echo "One"; sleep 1; echo "Two"; sleep 2; echo "Three"; sleep 3; echo "Done" ) &

You could also make this more script-looking:

(
echo "One"
sleep 1
echo "Two"
sleep 2
echo "Three"
sleep 3
echo "Done"
) &

Feel free to make use of the useful envar $BASH_SUBSHELL

( echo $BASH_SUBSHELL; ( echo $BASH_SUBSHELL ) )

Advertisement

Answer