在 Bash 中获取文件（仅当它存在时）的最简洁方法是什么？

Question

In Bash scripting, is there a single statement alternative for this?

if [ -f /path/to/some/file ]; then
    source /path/to/some/file
fi

The most important thing is that the filename is there only once, without making it a variable (which adds even more lines).

For example, in PHP you could do it like this

@include("/path/to/some/file"); // @ makes it ignore errors

Answer 1

Is defining your own version of @include an option?

include () {
    [[ -f "$1" ]] && source "$1"
}

include FILE

Answer 2

如果您担心不重复文件名的单行，也许：

FILE=/path/to/some/file && test -f $FILE && source $FILE

Answer 3

如果您担心警告（并且不存在源文件对您的脚本来说并不重要），只需摆脱警告：

source FILE 2> /dev/null

Answer 4

You could try

test -f $FILE && source $FILE

If test returns false, the second part of && is not evaluated

Answer 5

This is the shortest I could get (filename plus 20 characters):

F=/path/to/some/file;[ -f $F ] && . $F

It's equivalent to:

F=/path/to/some/file 
test -f $F && source $F

To improve readability, I prefer this form:

FILE=/path/to/some/file ; [ -f $FILE ] && . $FILE

Answer 6

If you want to always get a clean exit code, and continue no matter what, then you can do:

source ~/.bashrc || true && echo 'is always executed!'

And if you also want to get rid of the error message then:

source ~/.bashrc 2> /dev/null || true && echo 'is always executed!'

Answer 7

如果您不关心脚本的输出，您可以使用以下内容将标准错误重定向到/dev/null ：

$ source FILE 2> /dev/null