Last date of each month in a calender year [duplicate] - linux

Question

As we know that each year have the following max day in each month as follows:
Jan - 31 days
Feb - 28 days / 29 days (leap year)
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
How to I get bash to return the value (last day of each month) for the current year without using if else or switch or while loop?

Answer 1

my take:
for m in {1..12}; do
date -d "$m/1 + 1 month - 1 day" "+%b - %d days";
done
To explain: for the first iteration when m=1 the -d argument is "1/1 + 1 month - 1 day" and "1/1" is interpreted as Jan 1st. So Jan 1 + 1 month - 1 day is Jan 31. Next iteration "2/1" is Feb 1st, add a month subtract a day to get Feb 28 or 29. And so on.

Answer 2

cat <<EOF
Jan - 31 days
Feb - `date -d "yesterday 3/1" +"%d"` days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
EOF

Answer 3

cal $(date +"%m %Y") |
awk 'NF {DAYS = $NF}; END {print DAYS}'
This uses the standard cal utility to display the specified month, then runs a simple Awk script to pull out just the last day's number.

Answer 4

Assuming you allow "for", then the following in bash
for m in {1..12}; do
echo $(date -d $m/1/1 +%b) - $(date -d "$(($m%12+1))/1 - 1 days" +%d) days
done
produces this
Jan - 31 days
Feb - 29 days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
Note: I removed the need for cal
For those that enjoy trivia:
Number months from 1 to 12 and look at the binary representation in four
bits {b3,b2,b1,b0}. A month has 31 days if and only if b3 differs from b0.
All other months have 30 days except for February.
So with the exception of February this works:
for m in {1..12}; do
echo $(date -d $m/1/1 +%b) - $((30+($m>>3^$m&1))) days
done
Result:
Jan - 31 days
Feb - 30 days (wrong)
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days

Answer 5

Try using this code
date -d "-$(date +%d) days month" +%Y-%m-%d

Answer 6

Returns the number of days in the month compensating for February changes in leap years without looping or using an if statement
This code tests date to see if Feb 29th of the requested year is valid, if so then it updates the second character in the day offset string. The month argument selects the respective substring and adds the month difference to 28.
function daysin()
{
s="303232332323" # normal year
((!($2%4)&&($2%100||!($2%400)))) && s=313232332323 # leap year
echo $[ ${s:$[$1-1]:1} + 28 ]
}
daysin $1 $2 #daysin [1-12] [YYYY]

Answer 7

On a Mac which features BSD date you can just do:
for i in {2..12}; do date -v1d -v"$i"m -v-1d "+%d"; done
Quick Explanation
-v stands for adjust. We are adjusting the date to:
-v1d stands for first day of the month
-v"$i"m defined the month e.g. (-v2m for Feb)
-v-1d minus one day (so we're getting the last day of the previous month)
"+%d" print the day of the month
for i in {2..12}; do date -v1d -v"$i"m -v-1d "+%d"; done
31
28
31
30
31
30
31
31
30
31
30
You can add year of course. See examples in the manpage (link above).

Answer 8

Contents of script.sh:
#!/bin/bash
begin="-$(date +'%-m') + 2"
end="10+$begin"
for ((i=$begin; i<=$end; i++)); do
echo $(date -d "$i month -$(date +%d) days" | awk '{ printf "%s - %s days", $2, $3 }')
done
Results:
Jan - 31 days
Feb - 29 days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days

Answer 9

for m in $(seq 1 12); do cal $(date +"$m %Y") | grep -v "^$" |tail -1|grep -o "..$"; done
iterate from 1 to 12 (for...)
print calendar table for each month (cal...)
remove empty lines from output (grep -v...)
print last number in the table (tail...)
There is no sense, to avoid using cal, because it is required by POSIX, so should be there

Answer 10

A variation for the accepted answer to show the use of "yesterday"
$ for m in {1..12}; do date -d "yesterday $m/1 + 1 month" "+%b - %d days"; done
Jan - 31 days
Feb - 28 days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days
How it works?
Show the date of yesterday for the date "month/1" after adding 1 month

Answer 11

I needed this few times, so when in PHP comes with easy in bash is not,
so I used this till throw me error "invalid arithemtic operator"
and even with warrings in spellcheck ( "mt" stands for month, "yr" for year )
last=$(echo $(cal ${mt} ${yr}) | awk '{print $NF}')
so this works fine...
### get last day of month
#
# implement from PHP
# src: https://www.php.net/manual/en/function.cal-days-in-month.php
#
if [ $mt -eq 2 ];then
if [[ $(bc <<< "${yr} % 4") -gt 0 ]];then
last=28
else
if [[ $(bc <<< "${yr} % 100") -gt 0 ]];then
last=29
else
[[ $(bc <<< "${yr} % 400") -gt 0 ]] && last=28 || last=29
fi
fi
else
[[ $(bc <<< "(${mt}-1) % 7 % 2") -gt 0 ]] && last=30 || last=31
fi

Answer 12

Building on patm's answer using BSD date for macOS (patm's answer left out December):
for i in {1..12}; do date -v1m -v1d -v+"$i"m -v-1d "+%b - %d days"; done
Explanation:
-v, when using BSD date, means adjust date to:
-v1m means go to first month (January of current year).
-v1d means go to first day (so now we are in January 1).
-v+"$i"m means go to next month.
-v-1d means subtract one day. This gets the last day of the previous month.
"+%b - %d days" is whatever format you want the output to be in.
This will output all the months of the current year and the number of days in each month. The output below is for the as-of-now current year 2022:
Jan - 31 days
Feb - 28 days
Mar - 31 days
Apr - 30 days
May - 31 days
Jun - 30 days
Jul - 31 days
Aug - 31 days
Sep - 30 days
Oct - 31 days
Nov - 30 days
Dec - 31 days