Comparing dates in awk shell

Question

Hello I'm trying to make a script to search for specific info from a file and print it. My case is this : I have a file in the format of : id|lastname|firstname|birthday| . I want to call the script and given a date argument and the file to make it show me all the "people" born after the date I've given. Let me show you my code :

#!/bin/bash case $1 in --born-since ) d=($2 +%F); # this one puts the date I've given into the variable d grep -vE '^#' $4 | awk -F "|" ' $4 >= $d ' ;; esac

I call this script in the form of :

./script --born-since <date> -f <file>

Point is it's not doing what I want it to do. I prints wrong results. For example in a file with 4 dates ( 1989-12-03,1984-02-18,1988-10-14,1980-02-02), given the date of 1985-05-13 it prints only the person with date 1984-02-18 which is incorrect. It's probably comparing something else and not the date. Any advice ?

yeah my bad,it's wrong, it's just too many different results from different entries, therefore I got a bit confused, editing — Dimitris Delis
– Dimitris Delis, Commented Oct 14, 2017 at 17:38

RomanPerekhrest · Accepted Answer · 2017-10-14 17:44:59Z

1

With single awk process:

awk -v d="1985-09-09" -F'|' '$4 >= d' file

answered Oct 14, 2017 at 17:44

RomanPerekhrest

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2 Comments

Dimitris Delis Over a year ago

This works, although for dates equal it doesn't. Thank you though since I can try and experiment around that! Is it possible to compare for greater and less (for 2 different dates of course) in a single command?)

RomanPerekhrest Over a year ago

@DimitrisDelis, yes, it's possible

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