Question

我想查看文件列表并检查它们是否存在，如果文件不存在则发出错误并退出。我写了以下代码：

FILES=( file1.txt file2.txt file3.txt )

for file in ${FILES[@]}; do
        [ -e "${file}" ] || ( echo "ERROR: ${file} does not exist" >&2 && exit )
done

它运行时没有错误，并产生以下内容（如果没有文件存在）：

ERROR: file1.txt does not exist
ERROR: file2.txt does not exist
ERROR: file3.txt does not exist

为什么“退出”从未执行过？另外，我想知道我正在尝试做的事情的首选方式（用括号控制分组）。

Answer 1

可能因为您运行( echo "ERROR: ${file} does not exist" >&2 && exit )作为子进程（您的命令在inside（）中）？所以你要退出子流程。这是你的shell脚本的痕迹（我用set -x得到了它）：

+ FILES=(file1.txt file2.txt file3.txt)
+ for file in '${FILES[@]}'
+ '[' -e file1.txt ']'
+ echo 'ERROR: file1.txt does not exist'
ERROR: file1.txt does not exist
+ exit
+ for file in '${FILES[@]}'
+ '[' -e file2.txt ']'
+ echo 'ERROR: file2.txt does not exist'
ERROR: file2.txt does not exist
+ exit
+ for file in '${FILES[@]}'
+ '[' -e file3.txt ']'
+ echo 'ERROR: file3.txt does not exist'
ERROR: file3.txt does not exist
+ exit

这有效：

set -x
FILES=( file1.txt file2.txt file3.txt )

for file in ${FILES[@]}; do
        [ -e "${file}" ] || echo "ERROR: ${file} does not exist" >&2 && exit
done

将set -x放入您的档案中并亲眼看看。

或者像这样

set -x
FILES=( file1.txt file2.txt file3.txt )

for file in ${FILES[@]}; do
        [ -e "${file}" ] || (echo "ERROR: ${file} does not exist" >&2) && exit
done

<强>更新
我猜你问的是bash - grouping Commands

这是在同一过程中分组和执行

FILES=( file1.txt file2.txt file3.txt )

for file in ${FILES[@]}; do
        [ -e "${file}" ] || { echo "ERROR: ${file} does not exist" >&2; exit; }
done

这是在子流程

中进行分组和执行

FILES=( file1.txt file2.txt file3.txt )

for file in ${FILES[@]}; do
        [ -e "${file}" ] || ( echo "ERROR: ${file} does not exist" >&2; exit )
done

bash优先级与复合表达式

1 个答案: