使用dot运行shell脚本为什么“echo $ 0”print -bash

Question

use dot to run a shell script why does "echo $0" print -bash

Assuming I have a snippet of code below

#!/usr/bin/env bash
echo $0

I execute using dot (.)

. ./script.sh

Output:

-bash

However,

if i run without dot

Output is the path name combined with the script name as I expected.

Answer 1

获取脚本在现有解释器中运行它而不是exec新脚本，这意味着$0未正常初始化。

Answer 2

When you run a script with . , it's essentially as if you typed the contents of the script directly on your shell command line. So echo "$0" will output exactly what it would output if you just typed it.

The special variables like $0 are only initialized when you start up a new shell, which . doesn't do, or else it wouldn't be very useful.

However, if you're using bash version 3 or greater, you can always find out what file is currently being executed thus:

echo "$BASH_SOURCE"

which works no matter how the file is invoked.

Answer 3

. is an alias for source :

Don't use this command for run a script, it has been created for import source files.
You can see that here : Advanced Bash scripting guide