如何在 postgresql 中按周分组并按天区分

Question

Sample contents are:

id	created_dt	data
1	2023-01-14 11:52:41	{"customers": 1, "payments: 2}
2	2023-01-15 11:53:43	{"customers": 1, "payments: 2}
3	2023-01-18 11:51:45	{"customers": 1, "payments: 2}
4	2023-01-15 11:50:48	{"customers": 1, "payments: 2}

ID 4 or 2 should be distinct.

I want to get a result as follows:

year	week	customers	payments
2023	2	2	4
2023	3	1	2

I solved this problem in this way

SELECT
    date_part('year', sq.created_dt) AS year,
    date_part('week', sq.created_dt) AS week,
    sum((sq.data->'customers')::int) AS customers,
    sum((sq.data->'payments')::int) AS payments
FROM 
    (SELECT DISTINCT ON (created_dt::date) created_dt, data 
     FROM analytics) sq
GROUP BY 
    year, week
ORDER BY 
    year, week;

However, that subquery greatly complicates the query. Is there is a better method?

I need group the data by each week, however I also need to remove duplicate days.

Answer 1

Generate series to create the join table would solve the problem:

SELECT sum((sq.data->'customers')::int) as customers,
sum((sq.data->'payments')::int) as payments,
date_part('year', dategroup ) as year,
date_part('week', dategroup ) as week,
FROM generate_series(current_date , current_date+interval '1 month' , interval'1 week') AS dategroup
JOIN analytics AS a ON a.created_dt >= dategroup AND a.created_dt <= a.created_dt+interval '1 week'
GROUP BY dategroup
ORDER BY dategroup

Answer 2

First of all, I think your query is quite simple and understandable.

Here is the query with a with -query in it, in some point it adds more readabilty:

WITH unique_days_data AS (
  SELECT DISTINCT created_dt::date, data_json
  FROM analytics)
SELECT 
    date_part('year', ud.created_dt) as year,
    date_part('week', ud.created_dt) as week,
    sum((ud.data_json->'customers')::int) as customers,
    sum((ud.data_json->'payments')::int) as payments
FROM unique_days_data ud
GROUP BY year, week
ORDER BY year, week;

The difference is that the first query uses the DISTINCT clause, not the DISTINCT ON clause.

Here is the sql fiddle .

Answer 3

You can simplify it by adding partitioning on " created_id::date ", then filter last aggregated record for each week using FETCH FIRST n ROWS WITH TIES .

SELECT date_part('year', created_dt) AS year,
       date_part('week', created_dt) AS week,
       SUM((data->>'customers')::int) AS customers,
       SUM((data->>'payments')::int) AS payments
FROM analytics
GROUP BY year, week, created_dt::date
ORDER BY ROW_NUMBER() OVER(
             PARTITION BY date_part('week', created_dt) 
             ORDER     BY created_dt::date DESC
         )
FETCH FIRST 1 ROWS WITH TIES

Check the demo here .