mjhjmtu

#!/bin/bash
#!/usr/local/bin
#!/usr/sbin
#!/usr/bin
# Script to Check last working Day of the Month

echo " Enter Month and Year :"
read mon year
cal $mon $year| egrep "28|29|30|31"|awk 'BEGIN 
 var1=$NF;var2=NF;
 
 
 if (NF > 1 && NF < 7)
 val=$NF;
 else if (NF == 1)
 val=$NF-2;
 else if (NF == 7)
 val=$NF-1;
 
 
 print "Last Working Date is : " val;
 '

Script Output :

511@ubuntu:~/Unix$ ./test.sh
 Enter Month and Year :
4 2015
Last Working Date is : 30
511@ubuntu:~/Unix$ ./test.sh
 Enter Month and Year :
5 2015
Last Working Date is : 29
Last Working Date is : 29
511@ubuntu:~/Unix$ ./test.sh
 Enter Month and Year :
7 2015
Last Working Date is : 31
511@ubuntu:~/Unix$ ./test.sh
 Enter Month and Year :
1 2015
Last Working Date is : 30

Can someone please help why script print twice while we gives a input as below :

511@ubuntu:~/Unix$ ./test.sh
 Enter Month and Year :
5 2015
Last Working Date is : 29
Last Working Date is : 29

The problem is your egrep search is going to call awk once for every line that it sees a "28", "29", "30", or "31". Months where the 28th lands before the last calendar week will have awk called twice, since two lines match your search criteria

You want to always use the second line of the egrep search, so you can use a tail command to only see the last line:

#!/bin/bash
#!/usr/local/bin
#!/usr/sbin
#!/usr/bin
# Script to Check last working Day of the Month

echo " Enter Month and Year :"
read mon year
cal $mon $year| egrep "28|29|30|31"| tail -n 1 |awk 'BEGIN 
 var1=$NF;var2=NF;
 
 
 if (NF > 1 && NF < 7)
 val=$NF;
 else if (NF == 1)
 val=$NF-2;
 else if (NF == 7)
 val=$NF-1;
 
 
 print "Last Working Date is : " val;
 '

With date it is only a two line bash calculation:

#!/bin/bash

month="$1"
year="$2"

read -r dow day < <(date -d "$year/$month/1 +1 month -1 day" "+%u %d")

echo "Last working day of the month = $(( day - ( (dow>5)?(dow-5):0 ) ))"

The math is: if dow (the day of the week) is bigger than 5, subtract (dow-5) from the day, else, leave day unchanged.

Use as:

$ ./script 2 2015
Last working day of the month = 27

With ksh93:

$ printf "%(%A %B %d)Tn" "October 2018 last working day"
Wednesday October 31

Your script relies on the screen output of cal. To do things in a more proper way you could use the date command:

#!/bin/bash
# Calculates the last working day (Mon-Fri) of the month

echo "Enter Month and Year:"
read month year
lastdom=`date -d "$year/$month/1 + 1 month - 1 day" "+%d"`
dow=`date -d "$year/$month/1 + 1 month - 1 day" "+%u"`

if [ $dow -ge 6 ] 
then 
 lastwdom=$(($lastdom - $dow + 5))
else
 lastwdom=$(($lastdom))
fi

echo "Last working day is $lastwdom"

Your script repeats the print because awk is receiving two lines from egrep. But that has already been covered in other answers.

I want to explain some alternative way to solve the problem, shorter, easier.

The program cal could print the week starting on monday (which simplifies the math) when called as this cal -NMC month year. Using that:

#!/bin/bash

 lastday()
 printf 'Last Working Date of %s/%s = ' "$1" "$2";
 cal -NMC "$1" "$2" 

mon="$1"
year="$2"
lastday "$mon" "$year"

Description:

/[0-9]+/ Select lines with numbers (avoid empty line).

NF>5?5:NF Math: If more fields than 5 result is 5, else NF.

val=$( ... ) Select the value of the field.

END print val ' Only print the value of the last line (line with numbers).

Call it like this:

$ ./test.sh 4 2015
Last Working Date of 4/2015 = 30
$ ./test.sh 5 2015
Last Working Date of 5/2015 = 29
$ ./test.sh 7 2015
Last Working Date of 7/2015 = 31
$ ./test.sh 1 2015
Last Working Date of 1/2015 = 30
$ ./test.sh 9 2015
Last Working Date of 9/2015 = 30
$

read -p "Please enter date 'yyyymmdd':" dat
yy=`echo $dat|cut -c 1-4`
mm=`echo $dat|cut -c 5-6`

case $mm in

01|02|03)
mm=12
yy=`expr $yy - 1`
dd=`cal $mm $yy|awk -F " " 'print $6'|grep -v ^$|tail -1`
echo $yy$mm$dd
;;

04|05|06)
mm=03
dd=`cal $mm $yy|awk -F " " 'print $6'|grep -v ^$|tail -1`
echo $yy$mm$dd
;;

07|08|09)
mm=06
dd=`cal $mm $yy|awk -F " " 'print $6'|grep -v ^$|tail -1`
echo $yy$mm$dd
;;

10|11|12)
mm=09
dd=`cal $mm $yy|awk -F " " 'print $6'|grep -v ^$|tail -1`
echo $yy$mm$dd
;;

*) echo "Invalid month"
exit;;
esac