如何检查Bash中是否存在某些文件？

问题描述

在 bash 脚本中，我必须检查是否存在多个文件。

In a bash script, I have to check for the existence of several files.

我知道这样做很尴尬，如下所示，但这意味着我的主程序必须位于该丑陋的嵌套结构内：

I know an awkward way to do it, which is as follows, but that would mean that my main program has to be within that ugly nested structure:

if [ -f $FILE1 ]
then
if [ -f $FILE2 ]
then
  echo OK
  # MAIN PROGRAM HERE
fi
fi

以下版本不起作用：

([ -f $FILE1 ] && [ -f $FILE2 ]) ||  ( echo "NOT FOUND"; exit 1 )
echo OK

打印

NOT FOUND
OK

是否有一种优雅的方法来做到这一点？

Is there an elegant way to do this right?

更新：查看已接受的答案。另外，在优雅方面，我喜欢：

UPDATE: See the accepted answer. In addition, in terms of elegance I like Jonathan Leffler's answer:

arg0=$(basename $0 .sh)
error()
{
    echo "$arg0: $@" 1>&2
    exit 1
}

[ -f $FILE2 ] || error "$FILE2 not found"
[ -f $FILE1 ] || error "$FILE1 not found"

推荐答案

怎么样

if [[ ! ( -f $FILE1 && -f $FILE2 ) ]]; then
    echo NOT FOUND
    exit 1
fi

# do stuff
echo OK

请参见 help [[和 help test [[样式测试可用的选项。另请阅读。

See help [[ and help test for the options usable with the [[ style tests. Also read this faq entry.

您的版本不起作用，因为（...）会生成一个新的子shell，在其中执行 exit 。因此，它仅影响该子shell，而不影响执行脚本。

Your version does not work because (...) spawns a new sub-shell, in which the exit is executed. It therefor only affects that subshell, but not the executing script.

下面的方法代替了，在 {...}之间执行命令。 code>在当前shell中。

The following works instead, executing the commands between {...} in the current shell.

我还应该注意，您必须引用两个变量，以确保不会发生不必要的扩展或单词拆分（它们必须作为一个参数传递给 [）。

I should also note that you have to quote both variables to ensure there is no unwanted expansion or word splitting made (they have to be passed as one argument to [).

[ -f "$FILE1" ] && [ -f "$FILE2" ] || { echo "NOT FOUND"; exit 1; }

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