如何在postgresql中按周分组

Question

我有一个commits包含以下列的数据库表:

id | author_name | author_email | author_date(timestamp)| total_lines

样本内容如下:

1 | abc | abc@xyz.com | 2013-03-24 15:32:49 | 1234
2 | abc | abc@xyz.com | 2013-03-27 15:32:49 | 534
3 | abc | abc@xyz.com | 2014-05-24 15:32:49 | 2344
4 | abc | abc@xyz.com | 2014-05-28 15:32:49 | 7623

我想得到如下结果:

id | name | week | commits
1  | abc  | 1    | 2
2  | abc  | 2    | 0

我在网上搜索类似的解决方案但无法获得任何有用的解决方案.

我试过这个查询:

SELECT      date_part('week', author_date::date) AS weekly,
        COUNT(author_email)           
FROM        commits
GROUP BY    weekly
ORDER BY weekly

但它不是正确的结果.

Answer 1

如果你有多年,你也应该考虑这一年.一种方法是:

SELECT date_part('year', author_date::date) as year,
       date_part('week', author_date::date) AS weekly,
       COUNT(author_email)           
FROM commits
GROUP BY year, weekly
ORDER BY year, weekly;

一种更自然的方式来编写它date_trunc():

SELECT date_trunc('week', author_date::date) AS weekly,
       COUNT(author_email)           
FROM commits
GROUP BY weekly
ORDER BY weekly;

Answer 2

问这个问题已经很长时间了。
无论如何，如果有任何人通过这个。

如果你希望所有的计数中间周以及那里有没有提交/记录，你可以通过提供一个得到它start_date，并end_date以generate_series()功能

SELECT t1.year_week week, 
       t2.commit_count 
FROM   (SELECT week, 
               To_char(week, 'IYYY-IW') year_week 
        FROM   generate_series('2020-02-01 06:06:51.25+00'::DATE, 
               '2020-04-05 12:12:33.25+00':: 
               DATE, '1 week'::interval) AS week) t1 
       LEFT OUTER JOIN (SELECT To_char(author_date, 'IYYY-IW') year_week, 
                               COUNT(author_email)             commit_count 
                        FROM   commits 
                        GROUP  BY year_week) t2 
                    ON t1.year_week = t2.year_week; 

输出将是：

     week | commit_count  
----------+-------------
2020-05   | 2
2020-06   | NULL  
2020-07   | 1