通过bash查找特定文件的数量

Question

Fill in the dots on the next Unix command, like so, the standard out gives an overview per file type with the number of files in the /dev directory. In this overview, all filetypes must be listed in descending order of the number of found files of the certain type. If there are filetypes with an equal number of files, they must be listed in alphabetical order.

$ find /dev -ls | …
  7 c
  6 l
  3 d

Tips:

The part already given with the find-command, also finds hidden files in the directory.

With help of the cut-command, you can select a certain part of a line, the two most important options are -f and -d. The first one splits the lines in columns. By default, the tab-character is used. With the option -d you can specify a custom delimiter.

tr, sort and uniq might be useful.

What I have so far:

find /dev -ls | tr \\t " " | tr -s " " | cut -f3 -d ' ' | cut -c-1 | sort | uniq -c | sort -r

But this doesn't seem to work...

Thanks in advance.

Answer 1

我喜欢在这种情况下使用 awk 而不是 tr

 find /dev -ls | gawk '{ c=substr($3,1,1) ; x[c]++ } END { for(y in x) print x[y] " " y }' | sort -n