I am new to unix...

How to get all the saturdays of a specific year?

for a specific month, i tried as below..
cal 02 2006 | awk '{print $7}'
but it is not giving all saturdays....

can anyone help me with this?

Thanks in advance,
Sumi

Hi

Try the following command

cal 01 2006 | tr -s " " | sed -e '1,2d' | awk -F" " '{print $1}'



use the following in awk
$1 - For Sundays
$2 - For Mondays
$3 - For Tuesdays
and so on

Hope this helps

Regards
Bobby

i tried the command u sent...for this command,

cal 02 2006 | tr -s " " | sed -e '1,2d' | awk -F" " '{print $7}'

output should be:
4
11
18
25

but the output i am getting for the above command:
11
18
25

Read the date arithmetic article in our faq section. My datecalc script can do stuff like this:
The first argument is the day of the week (0 - 6) and the second argument is the year.
./script 6 2006 will give you all saturdays for the current year.

i am using datecalc function....after subtracting 6 from dt1, i am getting
2006 1 15

dt1=2006 01 27
datecalc -a $dt1 - 6

but i want the output as
2006 01 15

i searched the forum... if i get the "day" individually and perform operation, i can use typedef. but i am using the full date and subtracting 6 from it. how to get the desired output as above?

typeset -Z2 month day
datecalc -a $dt1 - 6 | read year month day
echo $year $month $day

I have one text file which has many date fields.

Date field is in "mm/dd/yyyy" format.

I want to replace every yyyy to yy format.

If i have one year value namely 2005, i can replace as

s/2005/05/g

Since i have many values, how to generalise the the above replacement ?
Through script, i need to grep and replace.

Thanks in advance.