I have a date in this format: "27 JUN 2011" and I want to convert it to 20110627
Is it possible to do in bash?
I have a date in this format: "27 JUN 2011" and I want to convert it to 20110627
Is it possible to do in bash?
If you would like a bash function that works both on Mac OS X and Linux:
#
# Convert one date format to another
#
# Usage: convert_date_format <input_format> <date> <output_format>
#
# Example: convert_date_format '%b %d %T %Y %Z' 'Dec 10 17:30:05 2017 GMT' '%Y-%m-%d'
convert_date_format() {
local INPUT_FORMAT="$1"
local INPUT_DATE="$2"
local OUTPUT_FORMAT="$3"
local UNAME=$(uname)
if [[ "$UNAME" == "Darwin" ]]; then
# Mac OS X
date -j -f "$INPUT_FORMAT" "$INPUT_DATE" +"$OUTPUT_FORMAT"
elif [[ "$UNAME" == "Linux" ]]; then
# Linux
date -d "$INPUT_DATE" +"$OUTPUT_FORMAT"
else
# Unsupported system
echo "Unsupported system"
fi
}
# Example: 'Dec 10 17:30:05 2017 GMT' => '2017-12-10'
convert_date_format '%b %d %T %Y %Z' 'Dec 10 17:30:05 2017 GMT' '%Y-%m-%d'
It's enough to do:
data=`date`
datatime=`date -d "${data}" '+%Y%m%d'`
echo $datatime
20190206
If you want to add also the time you can use in that way
data=`date`
datatime=`date -d "${data}" '+%Y%m%d %T'`
echo $data
Wed Feb 6 03:57:15 EST 2019
echo $datatime
20190206 03:57:15