How convert Date variable format in bash

Question

I have bash variable named date_export that have 2021-09-22 as a value. I want assign it to other bash variable named DATE_EXPORT but having other format 20210922 I tried :

DATE_EXPORT=$(date -d ${date_export} + '%Y%m%d')

But, it does not work, any help, please

date -d "$date_export" '+%Y%m%d' should work

anubhava
– anubhava

2021-12-01 15:30:09 +00:00
Commented Dec 1, 2021 at 15:30 — anubhava
– anubhava, Commented Dec 1, 2021 at 15:30

chepner · Accepted Answer · 2021-12-01 15:30:04Z

2

The output format is a single word starting with +, not two separate arguments + and the format.

DATE_EXPORT=$(date -d "$date_export" +"%Y%m%d")

answered Dec 1, 2021 at 15:30

chepner

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user1934428 · Accepted Answer · 2021-12-01 15:43:39Z

1

date_export=2021-09-22
DATE_EXPORT=${date_export//-/}
echo $DATE_EXPORT # prints 20210922

answered Dec 1, 2021 at 15:43

user1934428

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Comments

Léa Gris · Accepted Answer · 2021-12-01 17:50:21Z

0

Clearing the dashes using POSIX-shell grammar:

DATE_EXPORT=$(IFS=-; printf %s $date_export)

Or without spawning a sub-shell:

DATE_EXPORT=${date_export%-*}
DATE_EXPORT=${DATE_EXPORT%-*}${DATE_EXPORT#*-}${date_export##*-}

answered Dec 1, 2021 at 17:50

Léa Gris

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