Awk Sort by Today's date (not alphabetical) - awk

Question

I have the following lines in a file. The first columns look like this (these are dates and time):
May 29 23:14:39
Dec 20 19:45:15
Nov 3 13:15:19
Sep 8 10:34:15
Mar 9 18:39:20
Jan 17 19:34:59
I would like to use awk to sort it by today's date. For example today is November 03 (Nov 3). Tomorrow it will be November 04 (Nov 4). The dates will be there accordingly as the days change and the file changes along. Now I would like the first line to be the today's date all the time / always. Is that even possible to sort it out like this using either awk, sed and the like?
Alphabetical sort does not work because it does 1,2,3,4 and so on and if today is January 20 for example, my first line will be January 1 anyway (not January 20) because it will do alphabetical order / sort.
Would appreciate any help / suggestions / pointers. Many thanks in advance.
P.S. Let me edit this as asked by Cyrus.
Well, the following code actually works, but... sed '/reject/!d' file.txt | sort -r -k2'... It goes like this: November 1, November, 2, November 3, and then October 31. I guess it grabs numbers like so 1,2,3,31. If I could get it to solve this it would also work. Thanks.
My desired output is to sort it by date with the current today's date to be the first line in my file all the time. However, the code above would also work for me if I could get it to count November 1, November 2, November 3, November 4 (instead of November 1, November 2, November 3, October 31).
P.S.S. That's another edit as per Ed's Morton request.
The dates are all there. I do not need to add anything. The only requirement is for today's date line to be the first one in that file all the time and then sort it out backwards in descending order. For example today is November 05, 2022 and this is the first line in the file like this Nov 5 12:45:89. Then all the other lines are for November 4, 3, 2, 1. Then all the other lines are for October 31, 30, 29, 28 and so on. It is supposed to go backwards and it has to start from the current date, that is from "today" all the time. For example tomorrow the first line has to be Nov 6 and everything else backwards. Then the day after tomorrow the first line has to be Nov 7 and everything else backwards and so on and so forth. I do not need to truncate anything. I do not need to add anything. I do not need to delete anything. All the data has to stay there and go backwards starting from the "current" day, whether it is today, tomorrow, after tomorrow and so on. I hope it's clear enough. Thanks.

Answer 1

It sounds like this, using any POSIX compliant versions of the tools, might be what you want but without expected output in the question it's a guess:
$ cat tst.sh
#!/usr/bin/env bash
awk -v today="$(date +'%F')" '
BEGIN {
OFS = "\t"
split(today,d,/-/)
year = d[1]
}
{
mthNr = index(" JanFebMarAprMayJunJulAugSepOctNovDec",$1) / 3
date = sprintf("%04d-%02d-%02d", year, mthNr, $2)
}
date <= today {
print date, $3, $0
}
' "${#:--}" |
sort -rk1,2 |
cut -f3-
$ ./tst.sh file
Nov 3 13:15:19
Sep 8 10:34:15
May 29 23:14:39
Mar 9 18:39:20
Jan 17 19:34:59
The above uses a DSU approach to solve the problem. It's sorting on both date and time so for the same date the output is similarly ordered by the time that day.