[SOLVED] Bash script for printing days in a specific month (taking into account a leap year)

michaelk · 04-25-2012, 12:22 PM

Does this work?

Code:

if (( ($year % 400) == 0 )) || (( $year % 4 == 0 && $year % 100 !=0 )); then
  echo "This is a leap year"
else 
  echo "this is not a leap year"
fi

shayno90 · 04-25-2012, 02:09 PM

Quote:

Originally Posted by grail

hmmm ... can see a typo but not read the questions and answers

It would appear that the code in post #6 works just fine for me:

Code:

$ ./leap_year.sh
The current month is April
04 has 30 days

Is that the same way you are running the code?

Yes running the script from the script directory works as you run it but not the other way!

Code:

root@localhost:~/script# ./leaptest.sh 
The current month is April
04 has 30 days

but not this way

Code:

root@localhost:~/script# sh leaptest.sh 
The current month is April
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
leaptest.sh: 40: [[: not found
Programming error: 04 not in expected 01-12 range: 04

Any idea why?

shayno90 · 04-25-2012, 02:16 PM

Quote:

Originally Posted by michaelk

Does this work?

Code:

if (( ($year % 400) == 0 )) || (( $year % 4 == 0 && $year % 100 !=0 )); then
  echo "This is a leap year"
else 
  echo "this is not a leap year"
fi

Yes, the code works when run as:

Code:

root@localhost:~/script# ./lyear.sh 
The current month is April
04 has 30 days

but not when run as:

Code:

root@slocalhost:~/script# sh lyear.sh 
The current month is April
lyear.sh: 13: Syntax error: word unexpected (expecting ")")

I am confused now as I thought you had to specify the script was a bash script before running?!

catkin · 04-25-2012, 08:43 PM

When invoked as sh (which may or may not be a link to bash on your system), bash runs in POSIX mode and the full bash feature set is not available. Try bash 1year.sh

shayno90 · 04-26-2012, 05:11 AM

Quote:

Originally Posted by catkin

When invoked as sh (which may or may not be a link to bash on your system), bash runs in POSIX mode and the full bash feature set is not available. Try bash 1year.sh

Thanks I invoked the wrong bash shell when running the script!

I will add this to the script also to see if I can get user input in the form of one argument (enter a year in the terminal to check if it is a leap year.)

Code:

#!/bin/bash
# This script will test the user input to see if we're in a leap year or not.

year=`date +%Y`
getyear=("$1"(year))


if [ ! $getyear == `date +%Y` ]; then
  echo "Year must be in the format YYYY!"
  exit
fi

if [ $[$getyear % 400] -eq "0" ]; then
  echo "This is a leap year.  February has 29 days."
elif [ $[$getyear % 4] -eq 0 ]; then
        if [ $[$getyear % 100] -ne 0 ]; then
          echo "This is a leap year, February has 29 days."
        else
          echo "This is not a leap year.  February has 28 days."
        fi
else
  echo "This is not a leap year.  February has 28 days."
fi

How can I get the user input to read in the format of "date +%Y"?

shayno90 · 04-26-2012, 06:57 AM

Did it using the read function but would like to know is it possible to do so from adding an argument to the script itself? (i.e. ./lyear.sh 2000)

Code:

#!/bin/bash
# This script will test the user to see if we're in a leap year or not.

echo "Type the year that you want to check (4 digits), followed by [ENTER]:"

read year

if (( ("$year" % 400 == "0") )) || (( ("$year" % 4 == "0") && ("$year" % 100 != "0") )); then

    echo "$year is a leap year"
else
    echo "This is not a leap year" 
fi

catkin · 04-26-2012, 07:07 AM

Something like

Code:

if [[ $1 = '' ]]; then
    read -p 'Enter year > ' year_in
else
    year_in=$1
fi

For robustness your script could check that year_in is an integer before doing arithmetic on it -- and an unsigned integer for sanity.

grail · 04-26-2012, 07:19 AM

I am a bit lost at what you want?

Code:

year=`date +%Y`
getyear=("$1"(year))

How were these ever to be the same? Also the second causes an error when run at the command line, even if you turn $1 into just 1.

You have solved with the read option which takes the entered value and assigns it to the variable supplied. How about you take out the middle man and apply it to the variable yourself,
which would then be a parameter from the command line

shayno90 · 04-26-2012, 08:10 AM

Quote:

Originally Posted by grail

I am a bit lost at what you want?

Code:

year=`date +%Y`
getyear=("$1"(year))

How were these ever to be the same? Also the second causes an error when run at the command line, even if you turn $1 into just 1.

You have solved with the read option which takes the entered value and assigns it to the variable supplied. How about you take out the middle man and apply it to the variable yourself,
which would then be a parameter from the command line

I wasn't sure how to assign the argument from the command line (i.e., ./lyear 2000) and then ensure the argument (2000) is passed/formatted according to `date +%Y` (YYYY) (of course I don't want the result from date, just how to format it like date).

catkin · 04-26-2012, 08:26 AM

Quote:

Originally Posted by shayno90

I wasn't sure how to assign the argument from the command line (i.e., ./lyear 2000) and then ensure the argument (2000) is passed/formatted according to `date +%Y` (YYYY) (of course I don't want the result from date, just how to format it like date).

Both the argument passed from the command line and the date output are strings so no special formatting is required.

grail · 04-26-2012, 10:51 AM

Here is a possible way to check you are getting up to 4 and only 4 digits:

Code:

regex='^[1-9][0-9]{,3}$'

user_date=$1

while [[ ! $user_date =~ $regex ]]; do read -p "Invalid value, please try again :> " user_date; done

kieranpilot · 08-22-2018, 07:43 PM

As seen here :

YEAR=2020
MONTH=2

for DAY in $(DAYS=`cal $MONTH $YEAR | awk 'NF {DAYS = $NF}; END {print DAYS}'` && seq -f '%02G' $DAYS) ;do
DATE="$YEAR-$MONTH-$DAY"
echo $DATE
done

astrogeek · 08-23-2018, 02:35 AM

@kieranpilot - You have posted to a thread which has had no activity in six years. Although your information may be relevant it is best to start your own thread if you are experiencing a similar problem to attract attention of currently active members.

Additionally, advertising is not permitted in the forums, so your promotional post has been moderated. If you wish to advertise please visit the LQ Jobs Marketplace or contact us via the link on any page.

MadeInGermany · 08-23-2018, 05:50 PM

Let GNU date do the calculations:

Code:

currmonth=`date +%Y-%m`
days=`date -d "$currmonth-01 + 1 month - 1 day" +%d`
echo "The current month is $currmonth and has $days days"