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There is a need to get first Saturday date of current month in my script. I am able to get last Saturday date from below command

date -d 'last Saturday' +%y%m%d

But unable to get First Saturday date. Even tried date -d 'First Saturday' +%y%m%d but not getting expected result.

My requirement is for example - in current month (July) - First Saturday date (200704)

Can anyone help me in getting first Saturday date of current month

2

With ksh93:

$ ksh93 -c 'printf "%(%y%m%d)T\n" "first saturday"'
200704
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  • will this command also work in perl script ?
    – Sin15
    Jul 29 '20 at 15:38
  • 1
    @Sin15 of course you could call a shell command from perl, but if your requirement is to generate the date from within perl, you should ask a separate question. As-is, this question is tagged with bash, so Answerers are assuming you are calling it from a shell script. (Notice this answer specifying the ksh93 shell).
    – Jeff Schaller
    Jul 29 '20 at 16:01
2

bash

Indented commands are just for better understanding.

$ target_day_id=6 # return value of date +%w; Sunday is 0
$ year_month="$(date +%Y-%m)"
$     echo $year_month
2020-07
$ day_id="$(date --date="${year_month}-01" +%w)"
$     echo $day_id
3
$ # July, 1st 2020 was a Wednesday
$ sat1_date="${year_month}-0$(( ( (7+target_day_id-day_id) % 7 ) + 1 ))"
$     echo $sat1_date
2020-07-04
$     # only formatting from now on
$ sat1_date_short="${sat1_date:2}"
$     echo $sat1_date_short
20-07-04
$ sat1_date_short_digits_only="${sat1_date_short//-/}"
$     echo $sat1_date_short_digits_only 
200704
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  • could you please explain.... Also if in future there is requirement to replace saturday with sunday .... What value should i change ?
    – Sin15
    Jul 29 '20 at 15:41
  • @Sin15 I added the result after each command so that you can see what the commands do. I had to change the code a bit in order to make it flexible. You just set the desired target day at the beginning. Jul 29 '20 at 16:19
1

With GNU date and a shell loop:

day=1
until [ $(date +%A -d $(date +%Y-%m-$(printf '%02d' "$day"))) = "Saturday" ]
do 
  ((day++))
done
date +%y%m$(printf '%02d' "$day")

This initializes a "day" variable to 1 to indicate the first day of the month, then loops forward day by day until the name of that day is Saturday. GNU date is required in order to use the -d flag to use a timestamp different from "now". The test for the loop breaks down like this:

  • [ $(date +%A ...) = "Saturday" ] -- compare the output of date (asking it for the current locale's full weekday name) to "Saturday"
  • date ... -d $(date +%Y-%m-$(printf '%02d' "$day")) -- give that date command a date to display (don't use the current time) with -d; that date is generated from a second, inner date command. That inner date command prints the current time in %Y-%m-(day) format, where %Y is the current year, %m is the current month, and we pass in the loop-variable 'day'. The innermost printf statement simply ensures that the day field is zero-padded to two characters.

At the end, it prints out the current year and month in your desired format, using the calculated Saturday day.

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  • To avoid locale issues, I'd use $(date +%w -d ...) -eq 6n Jul 29 '20 at 16:16
  • I was torn about how to approach it -- using day-of-week is good; in the end I decided to hard-code "Saturday" from the question. I also saw a comment from them about possibly changing to "Sunday", so having an obvious string to change might be easier to adapt.
    – Jeff Schaller
    Jul 29 '20 at 16:27

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