Check the last saturday of the current month

user_9364372 0
Viewed 141

I am trying to run a shell script which checks whether today is the last Saturday of the current month and then it runs some script. For eg- 28th is the last Sat of this month. This is the code I am trying:

#!/bin/bash
DoDay=$(ncal -h | awk ' /Su/ {print $(NF-1)}')
Date=(date +%d)

if [[$Date == $DoDay]]
then 
    echo "Last Saturday"
else 
    echo "not last"
fi

While executing, it shows as :

ncal:command not found.

Can anybody please help here? Thanks in advance

awk shell unix
4 Answers

Could you please try following. I am running it as a command you could save it in a script and could run it from cron to check it on daily basis too.

cal $(date +%m) $(date +%Y) | 
tac | 
awk -v date=$(date +%d) '
  count==1{  exit  }
  NF==7{
    ++count
    if($(NF-1)==date){
      print "Today is last Saturday of month."
      exit
    }
    else{
      print "Today is NOT last Saturday of month."
    }
}'

For today's date(its Friday today on 6th Nov 2020) following is the output.

Today is NOT last Saturday of month.

Explanation: Adding detailed explanation for above.

cal $(date +%m) $(date +%Y) |                       ##Running cal command with passing current month and today date to it
tac |                                               ##Passing cal output to tac command to print it from bottom to top, since we want to read last lines only.
awk -v date=$(date +%d) '                           ##Passing previous output to awk as an Input and create date variable which has today date in it.
  count==1{  exit  }                                ##If count variable is 1 thne exiting from program.
  NF==7{                                            ##Checking condition if number of fields is 7 then do following.
    ++count                                         ##Increasing count with 1 here.
    if($(NF-1)==date){                              ##Checking condition if last field is date then do following.
      print "Today is last Saturday of month."      ##Printing that its last Saturday for this month.
      exit                                          ##exiting from program from here.
    }
    else{                                           ##Else part of above if condition here.
      print "Today is NOT last Saturday of month."  ##printing that its NOT last Saturday of this month.
    }
}'
user_58665800

If you are using GNU AWK you might harness its time functions. I would check it following way:

BEGIN{nowtime=systime();nowweekday=strftime("%u",nowtime);nowmonth=strftime("%m",nowtime);nextmonth=strftime("%m",nowtime+604800);if(nowweekday==6 && nowmonth!=nextmonth){print "Last saturday of month"}else{print "Not last saturday of month"}}

Explanation: firstly I get current time in terms of seconds since start of epoch, then I get current weekday as number (1-Monday, 7-Sunday), current month as number (1-12) and month of date 7 days later (604800 is number of seconds in 7 days). If today is Saturday and 7 days later will be different month this is last saturday, otherwise it is not, which I print accordingly

user_107859750

Some alternatives to get the last Saturday of the month:

  • ncal

    $ ncal | awk '$1 == "Sa" {print $NF}'
28

  • cal

    $ cal | awk 'NF == 7 {day = $7} END {print day}'
28

  • GNU date with some shell arithmetic: This uses nested date calls to get the day and day-of-week of the last day of this month, then iterates backwards until the day-of-week is Saturday.

    $ read day dow < <(date -d "$(date  '+%Y-%m-01 +1 month -1 day')" '+%d %w')
$ while ((dow != 6)); do ((day--)); ((dow = (dow - 1 + 7) % 7)); done
$ echo $day
28
user_75520

GNU date has some text parsing powers that simplyfies this:

if test `date +%w` -eq 6 && test "$(date +%m)" -ne "$(date -d 'next saturday' +%m)"; then
    echo "Last Saturday"
else
    echo "not last"
fi
user_9092760
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