Format date with a 3-char month name abbreviation using AWK-CodePudding

I have a datetime format as shown in the example below, which I want to convert to dd-mm-yyyy hh:mm:ss with AWK. How can I do this?

Current format:

3Jun2020 9:33:24; HG3456
7Jun2020 15:25:10; CH4747
10Jun2020 8:49:18; EU4821
12Jun2020 7:13:57; PP3478

Desired output:

03-06-2020 09:33:24; HG3456
07-06-2020 15:25:10; CH4747
10-06-2020 08:49:18; EU4821
12-06-2020 07:13:57; PP3478

CodePudding user response：

perl -MPOSIX -pe 's{\b\d{10}(?=\d{9}\b)}{
  strftime("%Y-%m-%d %T.", localtime $&)}ge

CodePudding user response：

I would use GNU AWK for this task following way, let file.txt content be

3Jun2020 9:33:24; HG3456
7Jun2020 15:25:10; CH4747
10Jun2020 8:49:18; EU4821
12Jun2020 7:13:57; PP3478

then

awk '{sub(/Jan/,"-01-",$1);sub(/Feb/,"-02-",$1);sub(/Mar/,"-03-",$1);sub(/Apr/,"-04-",$1);sub(/May/,"-05-",$1);sub(/Jun/,"-06-",$1);sub(/Jul/,"-07-",$1);sub(/Aug/,"-08-",$1);sub(/Sep/,"-09-",$1);sub(/Oct/,"-10-",$1);sub(/Nov/,"-11-",$1);sub(/Dec/,"-12-",$1);print}' file.txt

output

3-06-2020 9:33:24; HG3456
7-06-2020 15:25:10; CH4747
10-06-2020 8:49:18; EU4821
12-06-2020 7:13:57; PP3478

Explantion: replace Jan using -01-, Feb using -02-, Mar using -03- and so on, then print. Disclaimer: code might need adjusting if you use other locale.

(tested in GNU Awk 5.0.1)