如果 Bash 中存在包含/源脚本
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Include/Source script if it exists in Bash
提问by Bart van Heukelom
In Bash scripting, is there a single statement alternative for this?
在 Bash 脚本中，是否有一个单独的语句替代？
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).
最重要的是文件名只出现一次，没有使它成为变量（这会增加更多行）。
For example, in PHP you could do it like this
例如，在 PHP 中你可以这样做
@include("/path/to/some/file"); // @ makes it ignore errors
回答by chepner
Is defining your own version of @includean option?
是否定义了您自己@include的选项版本？
include () {
[[ -f "" ]] && source ""
}
include FILE
回答by Bryan Oakley
If you're concerned about a one-liner without repeating the filename, perhaps:
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
回答by mzet
If you are concerned about warning (and lack of existence of sourced file isn't critical for your script) just get rid of the warning:
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），请删除警告：
source FILE 2> /dev/null
回答by Attila
You could try
你可以试试
test -f $FILE && source $FILE
If testreturns false, the second part of &&is not evaluated
如果test返回 false，&&则不评估第二部分
回答by Jakob
This is the shortest I could get (filename plus 20 characters):
这是我能得到的最短的（文件名加 20 个字符）：
F=/path/to/some/file;[ -f $F ] && . $F
It's equivalent to:
它相当于：
F=/path/to/some/file
test -f $F && source $F
To improve readability, I prefer this form:
为了提高可读性，我更喜欢这种形式：
FILE=/path/to/some/file ; [ -f $FILE ] && . $FILE
回答by ayaz
If you are not concerned with the output of the script, you could just redirect standard error to /dev/nullwith something like this:
如果您不关心脚本的输出，您可以将标准错误重定向到/dev/null以下内容：
$ source FILE 2> /dev/null
$ source FILE 2> /dev/null
