Question

我写了一个查询来检索我每天有多少网站注册：

SELECT created, COUNT(id)
FROM signups
GROUP BY created
ORDER BY created desc

但是，这仅检索人们实际注册的天数。如果没有人在一天内注册，我想在那一天返回0。有没有办法使用SQL来执行此操作，还是必须使用PHP解析结果？

Answer 1

假设created由于缺乏信息而属于date类型。

Postgres提供了很棒的generate_series()来简化：

SELECT d.created, COUNT(s.id) AS ct
FROM  (
   SELECT generate_series(min(created)
                        , max(created), interval '1 day')::date AS created
   FROM   signups
   ) d
LEFT   JOIN signups s USING (created)
GROUP  BY 1
ORDER  BY 1 DESC;

这将自动从表中检索最小和最大日期，并在每天之间提供一行。

Answer 2

您可以使用NULLIF功能：

   SELECT created, NULLIF(COUNT(id), 0)
     FROM signups
 GROUP BY created
 ORDER BY created desc

文档：http://www.postgresql.org/docs/8.1/static/functions-conditional.html

Answer 3

您应该在数据库中创建一个日历表（或在查询中生成它）并与您的日历表一起加入，然后您将获得0空白日期

SELECT calendar.c_date, COUNT(signups.id)
FROM calendar
left join signups on calendar.c_date=signups.created

GROUP BY c_date
ORDER BY c_date desc

Here is a way to make a calendar date in PostgreSQL

Answer 4

您需要使用具有一系列日期并与之结合的日历表

select cal.created,coalesce(total) as total from calender_table as cal left join
(
SELECT created, COUNT(id) as total
FROM signups
GROUP BY created
) as source on cal.created=source.created
ORDER BY cal.created desc

检索行计数并在没有行时返回0

4 个答案: