find different beginnings of files with bash

Question

I have a directory with n files in it, all starting with a date in format yyyymmdd.

Example:

20210208_bla.txt
20210208_bla2.txt
20210209_bla.txt

I want know how many files of a certain date I have, so output should be like:

20210208 112
20210209 96
20210210 213
...

Or at least find the different beginnings of the actual files (=the different dates) in my folder.

Thanks

Picaud Vincent · Accepted Answer · 2021-02-08T14:35:47.010

2

A very simple solution would be to do something like:

ls | cut -f 1 -d _ | sort -n | uniq -c

With your example this gives:

 2 20210208
 1 20210209

Update: If you need to swap the two columns you can follow: https://stackoverflow.com/a/11967849/2001017

ls | cut -f 1 -d _ | sort -n | uniq -c | awk '{ t = $1; $1 = $2; $2 = t; print; }'

which prints:

20210208 2
20210209 1

edited Feb 08 '21 at 14:35

answered Feb 08 '21 at 14:25

Picaud Vincent

Good approach. The `awk` command for swapping could be simplified since we know that there are only two columns: `awk {print $1 " " $2}` – Socowi Feb 08 '21 at 14:40
@Socowi thanks for the remark, that is clearly a better solution! – Picaud Vincent Feb 08 '21 at 14:46
1

Peeeerfect thanks! :) – Hyperhyper Feb 08 '21 at 14:54

1 Answers1