当我获取脚本时终端关闭（在开始时使用点运行）

Question

Here is my code:

#!/bin/bash 
if [[ $1 = "" ]]; then
    exit 0
fi
array=($(cat $1))
let b=${#array[@]}-1
count=0
for i in {1..7}; do
    for j in {30..37}; do
        for n in {40..47}; do
            if [[ $count -gt $b ]]; then
                printf '\n'
                printf '\e[0m'
                exit 1
            fi
            printf '\e[%s;%s;%sm%-5s' "$i" "$j" "$n" "${array[$count]}"
            printf '\e[0m'
            let count=$count+1
        done
        printf '\n'
    done
done
#printf '\n'
printf '\e[0m'
exit 0

The problem is that when I start it like this

. color.sh arg

or without argument, it just closes. I know that the reason for that is exit . Is there any way correct my code so I could start a script with dot at start and terminal wouldn't close after execution? I don't want to start it like this: ./script

Answer 1

Replace all exit with return . return inside a sourced script will even work with exit codes:

$ . <(echo "echo before; return 0; echo after")
before
$ echo $?
0
$ . <(echo "echo before; return 7; echo after")
before
$ echo $?
7

Answer 2

When you use the dot to run a script you are "sourcing" it, which means the interpreter reads and executes all the commands in that script in the context of the current environment without spawning a subshell, as if you had typed each yourself.

That's why if you source it you can set variables in a script that will remain after it has run, whereas running it in a subshell would encapsulate them, and they would go away when the script ends.

Accordingly, if you source a script that hits an exit , it causes the calling environment to exit. Use return as Socowi suggested.