linux - 仅列出文件的公共(public)父目录

Question

我正在搜索一个文件，比如“file1.txt”，find 命令的输出如下所示。

/home/nicool/Desktop/file1.txt
/home/nicool/Desktop/dir1/file1.txt
/home/nicool/Desktop/dir1/dir2/file1.txt

在上面的例子中，我只想要公共(public)父目录，在上面的例子中是“/home/nicool/Desktop”。如何使用 bash 来实现？请帮助找到此类问题的通用解决方案。

Answer 1

此脚本在每次迭代中读取行并存储公共(public)前缀:

# read a line into the variable "prefix", split at slashes
IFS=/ read -a prefix

# while there are more lines, one after another read them into "next",
# also split at slashes
while IFS=/ read -a next; do
    new_prefix=()

    # for all indexes in prefix
    for ((i=0; i < "${#prefix[@]}"; ++i)); do
        # if the word in the new line matches the old one
        if [[ "${prefix[i]}" == "${next[i]}" ]]; then
            # then append to the new prefix
            new_prefix+=("${prefix[i]}")
        else
            # otherwise break out of the loop
            break
        fi
    done

    prefix=("${new_prefix[@]}")
done

# join an array
function join {
    # copied from: http://stackoverflow.com/a/17841619/416224
    local IFS="$1"
    shift
    echo "$*"
}

# join the common prefix array using slashes
join / "${prefix[@]}"

例子:

$ ./x.sh <<eof
/home/nicool/Desktop1/file1.txt
/home/nicool/Desktop2/dir1/file1.txt
/home/nicool/Desktop3/dir1/dir2/file1.txt
eof
/home/nicool