如何在PostgreSQL中按周分组 [英] How to group by week in postgresql
本文介绍了如何在PostgreSQL中按周分组的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我有一个数据库表 commits ，其中包含以下列：
I've a database table commits with the following columns:
id | author_name | author_email | author_date（时间戳）|
total_lines
id | author_name | author_email | author_date (timestamp) | total_lines
示例内容为：
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
我在网上搜索了类似的解决方案，但找不到任何有用的解决方案。
I searched online for similar solutions but couldnt get any helpful ones.
我尝试了以下查询：
SELECT date_part('week', author_date::date) AS weekly,
COUNT(author_email)
FROM commits
GROUP BY weekly
ORDER BY weekly
但这不是正确的结果。
But its not the right result.
推荐答案
如果您有多个年份，则也应考虑年份。一种方法是：
If you have multiple years, you should take the year into account as well. One way is:
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;
更自然的写法是使用 date_trunc（）：
SELECT date_trunc('week', author_date::date) AS weekly,
COUNT(author_email)
FROM commits
GROUP BY weekly
ORDER BY weekly;
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