Question

我有以下shell脚本。

SCRIPT1

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

SCRIPT2

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

我得到以下输出：

来自Script2的
foo
来自Script2的
foo1

预期输出

来自Script2的
foo
来自Script1的
foo1

我知道source命令在这里播放了spoilsport。有没有办法将控制权恢复到Script1或任何其他替代方法来实现这一目的？

Answer 1

在采购readonly之前，您可以使用foo1将script2定义为只读。

#!/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

执行script1后，您会看到：

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

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

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.

Answer 2

如果你希望脚本2的源代码只影响foo的执行，一种可能性就是把它放在子shell中：

( . script2 ; foo ; )
foo1

输出：

foo from Script2
foo1 from Script1

Answer 3

在声明source script2

之前将foo1移至

#!/bin/bash
foo() {
echo "foo from Script1"  
}
source script2     #Move this line to before declaration of foo1()
foo1() {                                     
echo "foo1 from Script1"
}
foo
foo1

在shell脚本中退出源命令

3 个答案: