生成系列以创建连接表将解决以下问题：

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

首先，我认为你的查询相当简单易懂。
以下是包含with-query的查询，在某些方面它增加了更多可读性：

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;

区别在于第一个查询使用DISTINCT子句，而不是DISTINCT ON子句。
这是sql fiddle。

您可以通过在"* created_id：：date *"上添加分区来简化它，然后使用FETCH FIRST n ROWS WITH TIES过滤每周的最后一个聚合记录。

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

检查here演示。