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
yes, one of the 'functions' is 'sed'
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.
The sed string to do that is worse than writing a shell function to do it IMO -
#!/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.
Jim,
that was not the desired input format.
given a sample input file 'mySampleFile.txt':
070829
890829
050829
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'
198200--12--29
I think it should be:
sed 's/\([^0].\)\(..\)\(..\)/19\1-\2-\3/g;s/^\(0.\)\(..\)\(..\)/20\1-\2-\3/g' mySampleFile.txt
Regards,
Tayyab
Thank you for your answers,
I manage it like that :
#!/bin/ksh
export a
format ()
{
yr=`expr substr $1 1 2`
month=`expr substr $1 5 2`
day=`expr substr $1 3 2`
if [[ $yr > "30" ]] ; then
echo "19""$yr-$month-$day"
else
echo "20""$yr-$month-$day"
fi
}
while read a
do
format $a >> TriDate.txt
done < $1
I have got another question , how can I remove the first line and the last line of a file ???
thank you thank you
Another question on the format() function
how does it come that the test :
[[ $yr > "30" ]]
works, but "30" is alphanumeric not numeric
I think tests on alphanumeric are : =, !=
and tests on numerics are : -eq, -ne, -gt, -ge, -lt, -le
?????????????????????????????????????
because yr can be 06, it's easier to use a string comparison. For me.