CODEWITHSUNDEEP
postgresql
ronak
ronak

Reputation: 1778

function to calculate aggregate sum count in postgresql

Is there a function that calculates the total count of the complete month like below? I am not sure if postgres. I am looking for the grand total value.

2012-08=# select date_trunc('day', time), count(distinct column) from table_name group by 1 order by 1;

     date_trunc      | count 
---------------------+-------
 2012-08-01 00:00:00 |    22
 2012-08-02 00:00:00 |    34
 2012-08-03 00:00:00 |    25
 2012-08-04 00:00:00 |    30
 2012-08-05 00:00:00 |    27
 2012-08-06 00:00:00 |    31
 2012-08-07 00:00:00 |    23
 2012-08-08 00:00:00 |    28
 2012-08-09 00:00:00 |    28
 2012-08-10 00:00:00 |    28
 2012-08-11 00:00:00 |    24
 2012-08-12 00:00:00 |    36
 2012-08-13 00:00:00 |    28
 2012-08-14 00:00:00 |    23
 2012-08-15 00:00:00 |    23
 2012-08-16 00:00:00 |    30
 2012-08-17 00:00:00 |    20
 2012-08-18 00:00:00 |    30
 2012-08-19 00:00:00 |    20
 2012-08-20 00:00:00 |    24
 2012-08-21 00:00:00 |    20
 2012-08-22 00:00:00 |    17
 2012-08-23 00:00:00 |    23
 2012-08-24 00:00:00 |    25
 2012-08-25 00:00:00 |    35
 2012-08-26 00:00:00 |    18
 2012-08-27 00:00:00 |    16
 2012-08-28 00:00:00 |    11
 2012-08-29 00:00:00 |    22
 2012-08-30 00:00:00 |    26
 2012-08-31 00:00:00 |    17
(31 rows)
--------------------------------
      Total          |    12345

Upvotes: 1

Views: 7344

Answers (2)

Craig Ringer
Craig Ringer

Reputation: 324375

As best I can guess from your question and comments you want sub-totals of the distinct counts by month. You can't do this with group by date_trunc('month',time) because that'll do a count(distinct column) that's distinct across all days.

For this you need a subquery or CTE:

WITH day_counts(day,day_col_count) AS (
  select date_trunc('day', time), count(distinct column)
  from table_name group by 1
)
SELECT 'Day', day, day_col_count
FROM day_counts
UNION ALL
SELECT 'Month', date_trunc('month', day), sum(day_col_count)
FROM day_counts
GROUP BY 2
ORDER BY 2;

My earlier guess before comments was: Group by month?

select date_trunc('month', time), count(distinct column)
from table_name
group by date_trunc('month', time)
order by time

Or are you trying to include running totals or subtotal lines? For running totals you need to use sum as a window function. Subtotals are just a pain, as SQL doesn't really lend its self to them; you need to UNION two queries then wrap them in an outer ORDER BY.

Upvotes: 4

Clodoaldo Neto
Clodoaldo Neto

Reputation: 125214

select
    date_trunc('day', time)::text as "date",
    count(distinct column) as count
from table_name
group by 1
union
select
    'Total',
    count(distinct column)
from table_name
group by 1, date_trunc('month', time)
order by "date" = 'Total', 1

Upvotes: 1

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