Linux bash如何在复制命令中使用通配符的结果作为文件名

Question

我正在编写一个Linux脚本，将文件从文件夹结构复制到一个文件夹中。我想使用一个不同的文件夹名作为文件名的前缀。

我现在的剧本是这样的。但是，我似乎无法找到使用通配符中的文件夹名作为文件名的方法；

for f in /usr/share/storage/*/log/myfile.log*; do cp "$f" /myhome/docs/log/myfile.log; done

我现有的文件夹结构/文件如下所示，我希望按以下方式复制文件；

>/usr/share/storage/100/log/myfile.log    -->    /myhome/docs/log/100.log
>/usr/share/storage/100/log/myfile.log.1  -->    /myhome/docs/log/100.log.1
>/usr/share/storage/102/log/myfile.log    -->    /myhome/docs/log/102.log
>/usr/share/storage/103/log/myfile.log    -->    /myhome/docs/log/103.log
>/usr/share/storage/103/log/myfile.log.1  -->    /myhome/docs/log/103.log.1
>/usr/share/storage/103/log/myfile.log.2  -->    /myhome/docs/log/103.log.2

Answer 1

您可以使用正则表达式匹配来提取所需的组件，但可能更容易更改为/usr/share/storage，以便所需组件始终是路径上的第一个组件。

一旦这样做，使用各种参数展开操作符提取要使用的路径和文件名的部分就很简单了。

cd /usr/share/storage
for f in */log/myfile.log*; do
    pfx=${f%%/*}  # 100, 102, etc
    dest=$(basename "$f")
    dest=$pfx.${dest#*.}
    cp -- "$f" /myhome/docs/log/"$pfx.${dest#*.}"
done

Answer 2

一个选项是将for循环包装在另一个循环中：

for d in /usr/share/storage/*; do
    dir="$(basename "$d")"

    for f in "$d"/log/myfile.log*; do
        file="$(basename "$f")"
        # test we found a file - glob might fail
        [ -f "$f" ] && cp "$f" /home/docs/log/"${dir}.${file}"
    done
done

Answer 3

for f in /usr/share/storage/*/log/myfile.log*; do cp "$f" "$(echo $f | sed -re 's%^/usr/share/storage/([^/]*)/log/myfile(\.log.*)$%/myhome/docs/log/\1\2%')"; done