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Date format conversion function

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# 1  
Old 08-30-2006
Date format conversion function


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

# 2  
Old 08-30-2006
yes, one of the 'functions' is 'sed'
# 3  
Old 08-30-2006
date format conversion

I can have either :

070829 --> I want 2007-08-29
890829 --> 1989-08-29
# 4  
Old 08-30-2006
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.
# 5  
Old 08-30-2006
The sed string to do that is worse than writing a shell function to do it IMO -
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"
             echo "20"$yr-$month-$day"

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

Edit per vgersh99 observation.

Last edited by jim mcnamara; 08-30-2006 at 05:48 PM..
# 6  
Old 08-30-2006
that was not the desired input format.

given a sample input file 'mySampleFile.txt':

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

# 7  
Old 08-31-2006
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:
echo "801229" | sed 's/\([^0].\)\(..\)\(..\)/19\1-\2-\3/g;s/\(0.\)\(..\)\(..\)/20\1-\2-\3/g'

I think it should be:
sed 's/\([^0].\)\(..\)\(..\)/19\1-\2-\3/g;s/^\(0.\)\(..\)\(..\)/20\1-\2-\3/g' mySampleFile.txt


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