linux - 从 shell 脚本中的 source 命令退出

Question

我有以下 shell 脚本。

脚本1

#!/bin/bash
foo() {
echo "foo from Script1"  
}
foo1() {
echo "foo1 from Script1"
}
source script2
foo
fool

脚本2

#!/bin/bash
foo() {
echo "foo from Script2"  
}
foo1() {
echo "foo1 from Script2"
}

我得到以下输出:

foo from Script2

foo1 from Script2

预期输出是

foo from Script2

foo1 from Script1

我知道 source 命令在这里玩弄恶作剧。有什么方法可以将控制权交还给 Script1 或任何其他替代方法来实现此目的？

Answer 1

您可以使用 readonly 将 foo1 定义为只读，然后再获取 script2。

#!/bin/bash
foo() {
echo "foo from Script1"  
}
foo1() {
echo "foo1 from Script1"
}
readonly -f foo1         # This would define foo1 as readonly
                         # and it wouldn't change upon sourcing script2
source script2
foo
foo1

在执行 script1 时，您会看到:

script2: line 7: foo1: readonly function
foo from Script2
foo1 from Script1

如果您想要错误(输出中的第一行不被打印)，将错误流重定向到 /dev/null，即通过说 bash script1 2> 调用您的脚本/dev/null.

---

文档:

$ help readonly
readonly: readonly [-aAf] [name[=value] ...] or readonly -p
    Mark shell variables as unchangeable.

    Mark each NAME as read-only; the values of these NAMEs may not be
    changed by subsequent assignment.  If VALUE is supplied, assign VALUE
    before marking as read-only.

    Options:
      -a    refer to indexed array variables
      -A    refer to associative array variables
      -f    refer to shell functions
      -p    display a list of all readonly variables and functions

    An argument of `--' disables further option processing.

    Exit Status:
    Returns success unless an invalid option is given or NAME is invalid.