CODEWITHSUNDEEP
awk
marvin
marvin

Reputation: 31

Get lines with dates less than current date

I have a file which has a date at the beginning of each line. I want to get all lines which has a date that is less than the current date.

Here is what I have,

When I used this:

awk '$NF < "2020-10-27"' expiry.txt >sample.txt

I was able to get a desired output

But when I substitute the date above "2020-10-27" to date today , I got no result

My script:

date=$(date '+%Y-%m-%d')
awk '$NF < "$date"' expiry.txt

sample logs in the file

2020-05-30: /tmp/abcd
2019-08-21: /tmp/abcde
2020-03-22: /tmp/abcdef

Thanks

Upvotes: 3

Views: 1164

Answers (2)

glenn jackman
glenn jackman

Reputation: 246807

The problem with your code: the shell will not substitute the $date variable if it's within single quotes. So you are comparing the date in column 1 against the literal string "$date". And because the ASCII value of $ is smaller than the ASCII value of 2, the $1 value is always lexically greater than the string "$date". Hence, no output.

If you have GNU awk, you can use the builtin time functions

gawk '
  BEGIN {today = strftime("%Y-%m-%d", systime())}
  $1 < today
' Input_file

Upvotes: 2

RavinderSingh13
RavinderSingh13

Reputation: 133518

Based on your shown samples and attempts, could you please try following.

awk -v dat="$(date '+%Y-%m-%d')" -F':' '$1<dat' Input_file

Multiple fixes in OP's attempts:

  • There was no proper value passed from shell variable to awk code.
  • Now coming to major issue, OP has used $NF which is last field of line but as per sample date is coming in very first field of lines, so changed it to $1(first field).
  • OP has used default field separator(space for awk), to catch the exact date and make it first field I have made field separator as : so that only date value is picked up for match with dat variable.

Upvotes: 3

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