How to generate a year-quarter date from the command line?

Question

I'm interested in outputting a representation of the current year-quarter, as well as the year-quarter for the previous month.

If today is 2012 January 1st, I'd like to get

2012q1

and

2011q4

as the respective outputs.

Answer 1

One (kinda ugly) solution, using BASH arithmetic evaluation and the GNU date command:

echo $(date +%Y)q$(( ($(date +%-m)-1)/3+1 ))
echo $(date -d "-1 month" +%Y)q$(( ($(date -d "-1 month" +%-m)-1)/3+1 ))

Note that the %-m prevents date from 0-padding, so this will still work for August and September.

Answer 2

All solutions that divide by four fail, for instance November:

% echo $(( 11/4+1 ))
3

The correct math would be:

$(( (m-1)/3 +1 ))

And as such, the quarter of current and previous month would be:

echo curr ${y}q$(((m-1)/3+1))
if [ $m = 1 ]; then
  echo prev $((y-1))q4
else
  echo prev ${y}q$(((m-2)/3+1))
fi

It's only twelve values to check, really…

% for m in {1..12}; do echo $m Q$(((m-1)/3+1)); done
1 Q1
2 Q1
3 Q1
4 Q2
5 Q2
6 Q2
7 Q3
8 Q3
9 Q3
10 Q4
11 Q4
12 Q4

Answer 3

Use my dateutils:

dconv 2012-01-01 -f '%Y%Q'
=>
  2012Q1

The %q and %Q flags are specific to dateutils, and return the quarter as number or in the form Q<NUMBER>.

Answer 4

Probably, there is no direct solution.

You could use awk to avoid so many back-ticks.

date +"%Y %m" | awk '{q=int($2/4)+1; printf("%sq%s\n", $1, q);}'
date +"%Y %m" | awk '{q=int($2/4);y=$1;if (q==0){q=4;y=y-1;}; printf("%sq%s\n", y, q);}'

A perl solution would be cleaner but perl and DateTime are an heavy prerequisite.

#!/usr/bin/perl

use DateTime;

my $today = DateTime->now;
print "today: " . $today->year . "q" . $today->quarter . "\n";

my $ago = DateTime->now->subtract( months=> 4);
print "some time ago: " . $ago->year . "q" . $ago->quarter . "\n"

Answer 5

Split the format with date, calculate with awk, format with printf:

date +"%Y %m" | awk '{printf ("%4dq%1d\n", $1, ($2/4)+1)}'

Just date and bash:

echo $(date +%Yq)$(($(date +%m)/4+1))

Answer 6

An alternative, more as a curiosity. If GNU awk is involved, date is not needed:

awk 'BEGIN{print strftime("%Y")"q"int((strftime("%-m")-1)/3)+1}'

Answer 7

Call date to retrieve the current year and month, and do the rest with arithmetic in the shell.

set $(date '+%Y %m');
this_quarter=${1}q$(($2 / 4 + 1))
if [ $2 -eq 1 ]; then
  last_month_quarter=$(($1 - 1))q4
else
  last_month_quarter=${1}q$((($2 - 1) / 4 + 1))
fi

Answer 8

Year-Quarter for this month

date +"%Yq$(expr $(expr $(date +%m) - 1) / 3 + 1)"

Year-Quarter for yester-month

date +"%Yq$(expr $(expr $(date -d '-1 month' +%m) - 1) / 3 + 1)"

Answer 9

Basic math for this quarter and last month's quarter:

y1=$(date +%Y)
m1=$(date +%m)
q1=$(( (m1 - 1) / 3 + 1))
y2=$(( y1 - (m1 == 1) ))
m2=$(( (m1 + 10) % 12 + 1 ))
q2=$(( (m2 - 1) / 3 + 1 ))
echo This Quarter: $((y1))q$q1
echo Last Month Quarter: $((y2))q$q2

The script uses the following parts:

1. $(unix-cmd) - evaluates a command in shell script
2. $((expr)) - evaluates a mathematical expression
3. remapping 1..12 -> 1..4 uses the following math (m-1)/3+1
4. evaluating the previous month uses modulo math

Answer 10

There is now the %q format to show this information.

From the coreutils-8.26 release log from November 30, 2016:

New Features
...
date now accepts the %q format to output the quarter of the year.

And yes it works!

$ date "+%q"
4
$ date "+%Yq%q"
2016q4

Answer 11

If you want the fiscal 13-week quarter that's based on the ISO week calendar, then the new handy %q isn't going to work, sadly. Here's a version of @manatwork's dateless solution with awk/strftime.

awk 'BEGIN{$x=int((strftime("%V")-1)/13)+1;print strftime("%G")"q"($x>4?4:$x)}'

The last little ternary bit handles the ISO leap-week years where the last quarter has 14 weeks.