sed is your friend
Something like:
*sed "s/Jan/01/;s/Feb/02/"*
On Mon, 17 Jan 2022 at 10:58, אורי <u...@speedy.net> wrote:
> Hi,
>
> I want to check mail.log for how many emails are sent every day. The
> format of my mail.log is something like this:
>
> Jan 17 08:49:23 www .... (the rest of the log)
>
> I'm running a command such as:
>
> cat mail.log* |fgrep "status=sent (250 Ok"|awk '{print $1" "$2}'|sort
> -n|uniq -c
>
> And I receive the number of emails sent every day. But the date doesn't
> contain the year, the months are sorted alphabetically and the line of Jan
> 17 comes before the line of Jan 2. I would like to sort the lines according
> to the date order such as in yyyy-mm-dd and with including the year. How do
> I do it?
>
> אורי
> u...@speedy.net
> _______________________________________________
> Linux-il mailing list
> Linux-il@cs.huji.ac.il
> http://mailman.cs.huji.ac.il/mailman/listinfo/linux-il
>
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