BASH order strings by the last “fields” (after “/” symbol")

Question

I'm looking for a method to sort line alphabetically by their last "field". So:

if my output is (maybe by a grep command):

mike/downloads.png
mike/public/system.png
mike/root/alphabet.png

the result should be:

root/alphabet.png
downloads.png
public/system.png

beacuse "alphabet" , "downloads" and "system" are order alphabetically.

should I firts cut and sort them with " cut -f2 -d"/" | sort " ? and then merge the rest of the path?

or there is an easier way?

Any helps will be appreciated.

Thanks

(example modified)

Answer 1

Sort具有-t参数来指定字段定界符，而-k具有指定要排序的字段，因此您可以编写：

sort -t/ -k 3 

Answer 2

Thank you all! I have finally found what i was looking for

first

awk -F'/' '{print $NF,$0}' 

then

sort 

and finally

sed -n 's/[^/]*\///p'

and the output will be

folder/file.png
file.png
folder/folder2/file.png

Answer 3

这指定第三个字段，其字段分隔符为/

sort -t'/' -k 3 

Answer 4

As the number of fields is dynamic you could append the last field to the start of the line before sorting and remove it after:

$ awk -F'/' '{print $NF,$0}' file | sort | awk '{print $2}'
mike/root/alphabet.png
mike/downloads.png
mike/public/system.png