如何从一个初始启动脚本生成多个bash后台脚本以登录终端的正确行？

Question

I have a bash script something like this:

# one.sh, two.sh and three.sh do not depend on anything,
# and can run simultaneously in any order,
# and the execution time for each is random

# four.sh can only run once three.sh has finished

one.sh &
ONE=$!

two.sh &
TWO=$!

three.sh &
THREE=$!

wait $THREE
four.sh

where one.sh, two.sh and three.sh look something like this:

echo -n "doing stuff..."
#some command or set of commands
if [ $? ]
then
    echo $RESULT_PASS
else
    echo $RESULT_FAIL
fi

the output i'm getting is something like this:

doing stuff1...done
doing stuff2...
doing stuff3...
doing stuff4...done
7263
doing stuff5...done
doing stuff6...9823
doing stuff7...done 
9283

because some tasks in one script are not complete before tasks in another script have started.

the output i'm looking for is something like this:

doing stuff1...done
doing stuff2...done
doing stuff3...8373
doing stuff4...done
doing stuff5...1234
doing stuff6...fail
doing stuff7...done

I would be grateful for some guidance.

Thanks!

Answer 1

I would use GNU Parallel to run scripts in parallel. It is very simple, intuitive and flexible. So, to run three scripts in parallel and keep the output in order and wait till all three have finished:

parallel -k ::: ./one.sh ./two.sh ./three.sh

Sample Output

doing stuff1...pass
doing stuff2...pass
doing stuff3...pass

Search for examples on Stack Overflow by typing [gnu-parallel] in the search box - include the square brackets. You can run as many as you want in parallel, tag the output lines, round-robin large files across multiple processes, remote login to other machines to spread processes across a network, halt all other scripts if one fails, or if 10% fail, or if any pass...

---

Or you can ask GNU Parallel to tag each line with the name of the script:

parallel -k --tag ::: ./one.sh ./two.sh ./three.sh
./one.sh    doing stuff1...pass
./two.sh    doing stuff2...pass
./three.sh  doing stuff3...pass

Answer 2

Than you need to use wait in the correct order. Keep in mind that you can't know for sure when a process is finished, so you can't echo text in your child process and expect it to be printed in order. You should handle the output in the parent script.

#! /bin/bash
PIDS=(0 0 0)
CMDS=("sleep 1" "sleep 2" "sleep 1 && exit 1")
X=0

IFS=$'\n'
for i in "${CMDS[@]}"; do 
    eval "$i &"
    PIDS[$X]=$!
    ((X++))
done

for ((i=0; i < $X; i++)); do
    wait ${PIDS[$i]}
    echo "Exit status is $?"
done
exit 0

I've created a similar script before which might be helpful.

Answer 3

If you want commands to run concurrently and have their output not interleaved, you'll need to work at it. You can try moving the cursor around the screen....but it' really better if you do not. Doing this will make it so that your tool can only be used interactively which will greatly limit its scope. However, if you choose to ignore that advice, you can try something like:

#!/bin/sh

output() {
        # Write a message on line $1 of the tty
        local mv=$(tput cup $1 0)
        local clear=$(tput el)
        printf "%s%s%s" "$mv" "$2" "$clear"

}

do_stuff() {
        i=0
        log() {
                output $1 "$( printf "doing stuff$1 %s" $2 )"
        }
        while sleep ${3-1} && test $i -lt $2; do
                log $1 "...$((i++))"
        done
        log $1 done
}

tput clear
do_stuff 1 5 .5 &
do_stuff 2 12 .2 &
do_stuff 3 16 .1
do_stuff 4 15 .2 &
wait
echo