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How to find second and fourth Monday of the month?

 
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# 15  
Old 06-30-2015
That's because the algorithm assigns day 7 to the second week already, which is the case for the second Monday of September and December, 2015. On top, the empty lines are due to invalid dates for months that dont have a 31., or 29., 30., 31. for February. And, DATE and dayofmonth hold identical values. Try the modified script:
Code:
YEAR=2015
for MONTH in $(seq -w 1 12) 
  do for DATE in $(seq -w 1 31)
    do dayofweek=$(date -d"${YEAR}${MONTH}${DATE}" +%w)
       [ $? == 0 ] || continue
       weekofmonth=$(( 1 + (10#$DATE - 1) / 7 ))
       if [ $dayofweek -ne 1 -o \( $weekofmonth -ne 2 -a $weekofmonth -ne 4 \) ]
         then continue
         else MONDAY_DATE=`date -d"${YEAR}${MONTH}${DATE}" +%d.%m.%Y`
              echo "$MONDAY_DATE"
       fi
    done
  done

# 16  
Old 06-30-2015
Quote:
Originally Posted by RudiC
That's because the algorithm assigns day 7 to the second week already, which is the case for the second Monday of September and December, 2015. On top, the empty lines are due to invalid dates for months that dont have a 31., or 29., 30., 31. for February. And, DATE and dayofmonth hold identical values. Try the modified script:
Code:
YEAR=2015
for MONTH in $(seq -w 1 12) 
  do for DATE in $(seq -w 1 31)
    do dayofweek=$(date -d"${YEAR}${MONTH}${DATE}" +%w)
       [ $? == 0 ] || continue
       weekofmonth=$(( 1 + (10#$DATE - 1) / 7 ))
       if [ $dayofweek -ne 1 -o \( $weekofmonth -ne 2 -a $weekofmonth -ne 4 \) ]
         then continue
         else MONDAY_DATE=`date -d"${YEAR}${MONTH}${DATE}" +%d.%m.%Y`
              echo "$MONDAY_DATE"
       fi
    done
  done

Thanks RudiC.
Can you please advise how are you coming up with the value 10 in below code? Basically, I want to understand the thought process while selecting the value, so the logic will be clear and will help in long run.
Code:
weekofmonth=$(( 1 + (10#$DATE - 1) / 7 ))

# 17  
Old 06-30-2015
man bash:
Quote:
Constants with a leading 0 are interpreted as octal numbers. A leading 0x or 0X denotes hexadecimal. Otherwise, numbers take the form [base#]n, where the optional base is a decimal number between 2 and 64 representing the arithmetic base, and n is a number in that base.
August and September will make the calculation fail unless the base is specified. I can't tell how far this applies to other shells...
This User Gave Thanks to RudiC For This Post:
Prathmesh
# 18  
Old 07-01-2015
Ah, yes. I hadn't thought of Monday 7th being the first week.

Try:-
Code:
((weekofmonth=1+($dayofmonth-1)/7))

Sorry about that,
Robin Smilie
This User Gave Thanks to rbatte1 For This Post:
Prathmesh
# 19  
Old 07-01-2015
Quote:
Originally Posted by RudiC
man bash: August and September will make the calculation fail unless the base is specified. I can't tell how far this applies to other shells...
Thanks. Script works well.

When searched more, I found out that instead of using %d option of date, if we use %e it works well.

Please find below modified script:
Code:
#!/bin/bash

YEAR=2015
for MONTH in $(seq -w 1 12)
do
        for DATE in $(seq -w 1 31)
        do
                dayofmonth=$(date -d"${YEAR}${MONTH}${DATE}" +%e 2>/dev/null)
		[ $? == 0 ] || continue
                dayofweek=$(date -d"${YEAR}${MONTH}${DATE}" +%w)
                weekofmonth=$((($dayofmonth-1)/7+1))

                if [ $dayofweek -ne 1 -o \( $weekofmonth -ne 2 -a $weekofmonth -ne 4 \) ]
                then
                        #printf "Not running today.\n"
                        continue
                else
                        MONDAY_DATE=`date -d"${YEAR}${MONTH}${DATE}" +%d.%m.%Y`
                        echo "$MONDAY_DATE"
                fi
        done
done

Output:
Code:
$ sh Monday_Date.sh
12.01.2015
26.01.2015
09.02.2015
23.02.2015
09.03.2015
23.03.2015
13.04.2015
27.04.2015
11.05.2015
25.05.2015
08.06.2015
22.06.2015
13.07.2015
27.07.2015
10.08.2015
24.08.2015
14.09.2015
28.09.2015
12.10.2015
26.10.2015
09.11.2015
23.11.2015
14.12.2015
28.12.2015
$

---------- Post updated at 04:24 PM ---------- Previous update was at 04:21 PM ----------

Quote:
Originally Posted by rbatte1
Ah, yes. I hadn't thought of Monday 7th being the first week.

Try:-
Code:
((weekofmonth=1+($dayofmonth-1)/7))

Sorry about that,
Robin Smilie
Thanks Robin/RudiC and others for all your help.
Last edited by Prathmesh; 07-01-2015 at 07:52 AM.. Reason: editing content
 
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