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
Maybe something changed since 2011 but this worked for me:
$ date +"%Y%m%d"
20150330
No need for the -d
to get the same appearing result.
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'
Just with bash:
convert_date () {
local months=( JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC )
local i
for (( i=0; i<11; i++ )); do
[[ $2 = ${months[$i]} ]] && break
done
printf "%4d%02d%02d\n" $3 $(( i+1 )) $1
}
And invoke it like this
d=$( convert_date 27 JUN 2011 )
Or if the "old" date string is stored in a variable
d_old="27 JUN 2011"
d=$( convert_date $d_old ) # not quoted
date -d "25 JUN 2011" +%Y%m%d
outputs
20110625
On OSX, I'm using -f to specify the input format, -j to not attempt to set any date, and an output format specifier. For example:
$ date -j -f "%m/%d/%y %H:%M:%S %p" "8/22/15 8:15:00 am" +"%m%d%y"
082215
Your example:
$ date -j -f "%d %b %Y" "27 JUN 2011" +%Y%m%d
20110627
Source: Stackoverflow.com