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Change Date from one format to other


 
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# 1  
Old 09-10-2013
[Solved] Change Date from one format to other

Hi

I wish to change date from one format to another in unix.

eg: INPUT DATE: 2013159 (YEAR & NUMBER OF DAY)
OUTPUT DATE required: 20130608 (YYYYMMDD)

how to do it ?

Thanks in advance.
# 2  
Old 09-10-2013
Providing you have the GNU version of the date command available this should work.

Code:
#!/bin/bash
#
#

# check the command line
if [ $# -ne 1 ]
then
    echo "Usage: ${0##*/} <year and day of year> (i.e. 2013159)"
    exit 1
fi

# get the given year
yr=$(echo $1 | cut -c1-4)

# get the given day
dy=$(echo $1 | cut -c5-7)

# perform the conversion using the date command
output=$(date -d "${yr}-01-01 +$(( ${dy} - 1 ))days" +%Y%m%d)

# display results
echo "$output"

# done
exit 0

day2date 2013159
20130608

This User Gave Thanks to in2nix4life For This Post:
# 3  
Old 09-10-2013
Thanks a lot.
It worked beautifully Smilie
# 4  
Old 09-10-2013
No problem. Smilie
# 5  
Old 09-10-2013
If you don't have GNU date available, you can try this shell snippet:
Code:
MONTHS=(31 28 31 30 31 30 31 31 30 31 30 31)
DATE=2013159
YEAR=${DATE:0:4}
DAY=${DATE:4}
((MONTHS[1]+=YEAR%4==0))
while ((DAY > (sum+=MONTHS[i++]))) ; do : ; done; printf "%d%02d%02d\n" $YEAR $i $((DAY+MONTHS[$i-1]-sum)) 
20130608

# 6  
Old 09-10-2013
Hi.

See also dateutils for a number of date-specific utilities, such as:
Code:
dconv
       Convert DATE/TIMEs between calendrical systems.  If DATE/TIME is
       omitted date/times are read from stdin.


Best wishes ... cheers, drl
# 7  
Old 01-13-2014
Using dconv from dateutils:
Code:
dconv -i '%Y%D' 2013159
=> 2013-06-08

 

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