每当日期格式约束与日期格式列不匹配时，如何使用 awk 按日期列过滤 csv 文件？

Question

I need to filter a csv file whom 25th column has the following date format "YYYY-MM-DD hh:mm:ss", in order to only display the entire lines containing today's date, and put them in a separate file (hope that's clear).

I went over a few topics and tried to adapt solutions to my script but I can't make it work.

So right now, I am back to the following :

today= date '+%Y-%m-%d'

awk -F ";" '{ if ($25 == '$today') print }' input.csv > today.csv'

I am aware that I am asking something stupid like "if YYYY-MM-DD == YYYY-MM-DD hh:mm:ss, then do something"

My question is, is there a way to put a wildcard somewhere in my today's variable, or should I set two different variables like :

today_start = date '+%Y-%m-%d' 00:00:00

today_end = date '+%Y-%m-%d' 23:59:59

and then create a condition stating that if column value is between today_start and _today_end, something needs to be done.

Also, I need to keep the format of date column as it is (YYYY-MM-DD hh:mm:ss).

Thanks a lot in advance, hope someone can help !

Answer 1

You can use a regex to match the start of your field, ie match the first 10 characters (YYYY-MM-DD) of the field.

today=$(date '+%Y-%m-%d')
awk -v regex="^$today" -F';' '$25 ~ regex' input.csv > today.csv

This passes the value of the $today variable with -v to awk and prepends a ^ to match the start of the field.