I need to loop a perl script backwards in time through the dates of all Thursdays for the past few years.

I can easily get the date of last Thursday with
`date -d 'last Thursday'`

How can I get the date of the Thursday before that and so on?

I can't change the date of the System.

For calculations like this, change to Unix timestamp, or seconds.

First get the TS of last Thursday in seconds:
last_th=$(date -d "last Thursday" "+%s")
When you have that, substract 7 * 86400 seconds from it
before_last_th=$last_th-7*86400
Get the date of that day:
date -d "1970-01-01 +$before_last_th seconds" "+%F %Ts"

BTW, date is a shell command, does it exist in Perl as well? The substraction is pseudo code, adjust to your language.

jlinkels

I'm not sure if this is standard, but in my perl environment I can run shell commands like:
@array_of_directory_contents = `ls .`;

With the ` ticks.

Thank you for the info, I had the idea of doing it all in epoch time, but it seemed like a hassle.

Perl does have its date/time functions, don't know if you can get something fancy like "last thursday" without complicated loops. Using the shell command seems a good idea.

Whatever, for calculations like this, using epoch time is definitely not a hassle. I wrote a complicated project in PHP which involved nothing but schedules and time calculations. I started out using datetime formats, but later I used used exclusively epoch time in calculations. Only at the presentation end I did a conversion to or from datetime.

jlinkels

This should give you a head start.
If you want years before 1970 try this.

yr=$(( 1969+28 ))
subdate=1969
yr=$(( $subdate+28 ))
mo=0$(( 1 ))
da=0$(( 2 ))

date -d $yr$mo$da +%u' '%a' '%b' '%d' '$subdate
4 Thu Jan 02 1969

The solar cycle is usually every 28 years.
Well, you know what to do when days amd months are larger than 9.
heh
a=$( echo $( date -d $yr$mo$da +%u ))
if [ $a == 4 ]
then
echo $( date -d $yr$mo$da +%u' '%a' '%b' '%d' '$subdate )
fi

hth