Question

我有以下查询，我需要与同一天具有相同golddate的用户数量的计数，但是由于我必须使用GROUP BY，当那天没有人时它不会返回零值你能帮忙吗？

SELECT count( * ) AS total, DATE( `golddate` ) AS gold_date
                FROM `user`
                WHERE YEAR( `golddate` ) >=2014
                GROUP BY DATE( `golddate` )
                ORDER BY DATE( `golddate` ) ASC

我想展示类似

的内容

goldate | total

2012-12-10 | 23
2012-10-12 | 0

但问题是它永远不会返回零值日期，因为

分组

我试过以下但没有用，

SELECT
  u1.golddate
  -- this will count the number of links to each word
  -- if there are no links the COUNT() call will return 0
  , COUNT(u2.golddate) AS linkCount
FROM user u2
  LEFT JOIN user u1
    ON u1.user_id = u2.user_id
      OR u2.user_id = u1.user_id
GROUP BY u1.golddate

这甚至不起作用

SELECT COALESCE(COUNT(`golddate`), 0) AS total, DATE( `golddate` ) AS gold_count
                FROM `user`
                WHERE YEAR( `golddate` ) >=2014
                GROUP BY DATE( `golddate` )
                ORDER BY DATE( `golddate` ) ASC

Answer 1

正如其他人所提到的，您需要人为地创建缺失数据的日期。

使用RedFilter's cross join date generation code here，您可以在此处创建给定范围内的所有日期，然后将LEFT JOIN返回到您的实际数据。 COUNT（*）需要更改为user表中的某些内容，以便不计算虚拟日期。

SELECT count(u.golddate) AS total, a.Date AS golddate
FROM 
(
    select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date
    from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
    cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
    cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
) a  

LEFT OUTER JOIN `user` u
  ON DATE(u.golddate) = a.Date

WHERE a.Date between '2013-01-01' and '2013-01-10'

GROUP BY a.Date
ORDER BY a.Date ASC;

SqlFiddle here

MySQL GROUP BY计数中的零值

1 个答案: