文件 Shell 脚本中的日期比较

Question

[I have a file which has second column date in format mm/dd/yyyy.

ASVC038M8  09/10/2019
ASVC066M8  11/04/2020
ASVC075M8  11/07/2024
ASVC068M8  11/08/2020
ASVC069M8  11/13/2020
ASVC070M8  11/14/2020
ASVC047M8  12/08/2019
ASVC051M8  12/08/2019
ASVC037M8  12/21/2019
ASVC052M8  12/21/2019
ASVC043M8  12/28/2019
ASVC040M8  12/28/2019

I want to get information for those entries which have date more than 35 Days ago.. My current code is

Date=$(date +"%m/%d/%y" -d "+35 days")

cat test.txt | awk -v date="$Date" '$2  > date { print $1"  "$2 }')

In this, I am getting the output as

ASVC043M8  12/28/2019
ASVC040M8  12/28/2019

Not sure why I am not getting all the entries as there are few more entries which have a date more than 35 days ago.

Help, please.

thanks in advance.

Answer 1

Could you please try following(date command should be GNU date )

Date=$(date +"%Y-%m-%d" -d "+35 days")
awk -v date="$Date" '(substr($2,7)"-"substr($2,1,2)"-"substr($2,4,2)) > date' Input_file

OR as per Ed sir's suggestion in comments with split try:

Date=$(date +"%Y-%m-%d" -d "+35 days")
awk -v date="$Date" '{split($2,array,"/")} (array[3]"-"array[1]"-"array[2]) > date' Input_file

Answer 2

You can use awk mktime to convert dates into sortable dates, or use bash string processing. Using awk will be faster - important for large files.

awk -v from=$(date -d '35 days ago' +'%Y-%m-%d') '
{ dt = substr($2, 7) "-" substr($2, 1, 2) "-" substr($2, 4, 2) ;
  if ( dt < from ) print
}'