CODEWITHSUNDEEP
stringshelldatedate-formatting
Fabrício Pereira
Fabrício Pereira

Reputation: 1640

How to get YEAR info from simple date string (YYYYMMDD) with SHELL SCRIPT?

When I run this command to get year info from date '20181102' (format: YYYYMMDD), it works fine:

$ echo $(date --date="20181102" +"%Y")
2018

But when I try with this other date '20181104', I got an error:

$ echo $(date --date="20181104" +"%Y")
date: invalid date ‘20181104’

Is this any bug? Or did I do something wrong?

Upvotes: 0

Views: 1991

Answers (1)

KamilCuk
KamilCuk

Reputation: 141563

GNU date utility is a little crude and keeps on misunderstanding the input string. For a good measure, try to always give date dates in @seconds since epoch or in one forms as in man date like "2004-02-29 16:21:42". (Unless you have freebsd date which has -f option.)

str=20181102
date --date="$(echo "$str" | cut -c 1-4)-$(echo "$str" | cut -c 5,6)-$(echo "$str" | cut -c 7,8) 00:00:00" +%Y

But really if you know that first 4 characters are year, why don't you just get first 4 characters?

str=20181102
echo "$str" | cut -c 1-4

Also note that doing echo $(date ...) is pointless. It's just date .... The echo is not needed in any way.

Upvotes: 1

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