Date format conversion function - UNIX for Dummies Questions & Answers

Cecile · #1 08-30-2006

Hello,

does somebody knows about a function that would convert a date like:

YYMMDD into a date like YYYY-MM-DD ?

Thank you for your ideas

vgersh99 · #2 08-30-2006

yes, one of the 'functions' is 'sed'

Cecile · #3 08-30-2006

I can have either :

070829 --> I want 2007-08-29
890829 --> 1989-08-29

vgersh99 · #4 08-30-2006

Quote:

Originally Posted by Cecile

I can have either :

070829 --> I want 2007-08-29
890829 --> 1989-08-29

ok, the 'function' 'sed' should be able to do it for you.

jim mcnamara · #5 08-30-2006

The sed string to do that is worse than writing a shell function to do it IMO -

Code:

#!/bin/ksh
format ()
{
        yr=`expr substr $1 1 2`      
        month=`expr substr $1 5 2`
        day=`expr substr $1 7 2`
        if [[ $yr > "30" ]] ; then
             echo "19""$yr-$month-$day"
        else
             echo "20"$yr-$month-$day"
        fi
}

new_date=$(format "041012")
echo $new_date

Edit per vgersh99 observation.

vgersh99 · #6 08-30-2006

Jim,
that was not the desired input format.

given a sample input file 'mySampleFile.txt':

Code:

070829
890829
050829

Code:

sed 's/\([^0].\)\(..\)\(..\)/19\1-\2-\3/g;s/\(0.\)\(..\)\(..\)/20\1-\2-\3/g' mySampleFile.txt

tayyabq8 · #7 08-30-2006

Quote:

sed 's/\([^0].\)\(..\)\(..\)/19\1-\2-\3/g;s/\(0.\)\(..\)\(..\)/20\1-\2-\3/g' mySampleFile.txt

Script will fail on where year is 90, 80, 70 and so on:

Code:

echo "801229" | sed 's/\([^0].\)\(..\)\(..\)/19\1-\2-\3/g;s/\(0.\)\(..\)\(..\)/20\1-\2-\3/g'
198200--12--29

I think it should be:

Code:

sed 's/\([^0].\)\(..\)\(..\)/19\1-\2-\3/g;s/^\(0.\)\(..\)\(..\)/20\1-\2-\3/g' mySampleFile.txt

Regards,
Tayyab

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