Change the date format from text to number using single awk command

Question

I have extremely large log file that I need to process and save for csv format every hour.

the format is like below ( this line also after removed many characters using sed)

Apr-05 11:10:12 xxxx xxx xxxx xxxx xxxx

I need to save this as

04-05 11:10:12,xxxx,xxx,xxxx,xxxx,xxxx

I tried set variable using $1 and try to convert it using GNU date but not working

awk -v fixdate="($1 +%m-)" '{print fixdate"-"$2" "$3 "," $x "," $y "," $z}'

I prefer single line method to do this.

How can I assign $1 (Apr-05) part to awk variable and process it with GNU date before printing.

This might help: [Formatting date strings in a file with linux bash shell](https://stackoverflow.com/q/61015078/3776858) — Cyrus, Apr 17 '20 at 07:56

score 2 · Accepted Answer · answered Apr 17 '20 at 17:10

Depending on whether or not any of those xs can be a -, this might be all you need:

$ awk -F'-' '{
    gsub(/ /,",",$2); sub(/,/," ",$2);
    printf "%02d-%s\n", (index("JanFebMarAprMayJunJulAugSepOctNovDec",$1)+2)/3, $2
}' file
04-05 11:10:12,xxxx,xxx,xxxx,xxxx,xxxx

Change the date format from text to number using single awk command

1 Answers1