MySQL - 计算一周中每一天但每个发件人的所有行

Posted 2023-03-24

【中文标题】MySQL - 计算一周中每一天但每个发件人的所有行

【英文标题】：MySQL - Count all rows for each day of week but for each sender

【发布时间】：2015-07-21 14:32:52

【问题描述】：

我有这个查询，它输出每天的行数，如果没有行，它输出 0。

我现在有一个额外的字段要添加到名为“发件人”的查询中。我需要对每个发件人执行完全相同的查询。

如何执行查询，以便每个发件人在一周中的每一天获取值？

SELECT DAYNAME(DATE_SUB(CURDATE(), INTERVAL Days.n DAY)) AS `day`,
       COUNT(r.List_Date) AS `total`
FROM (SELECT 1 as n UNION ALL SELECT 2 as n UNION ALL 
      SELECT 3 as n UNION ALL SELECT 4 as n UNION ALL
      SELECT 5 as n UNION ALL SELECT 6 as n UNION ALL
      SELECT 7 as n
     ) Days LEFT JOIN
     returns r
     ON r.List_Date >= DATE_SUB(CURDATE(), INTERVAL Days.n DAY)
GROUP BY Days.n
ORDER BY Days.n DESC

【问题讨论】：

group by sender, days.n

我试过了，但没用

【参考方案1】：

您需要cross join 来获取所有行（每个发件人和一周中的每一天）。然后使用left join：

SELECT s.sender, DAYNAME(DATE_SUB(CURDATE(), INTERVAL Days.n DAY)) AS `day`,
       COUNT(r.List_Date) AS `total`
FROM (SELECT 1 as n UNION ALL SELECT 2 as n UNION ALL 
      SELECT 3 as n UNION ALL SELECT 4 as n UNION ALL
      SELECT 5 as n UNION ALL SELECT 6 as n UNION ALL
      SELECT 7 as n
     ) Days CROSS JOIN
     (SELECT DISTINCT sender FROM returns) s LEFT JOIN
     returns r
     ON r.List_Date >= DATE_SUB(CURDATE(), INTERVAL Days.n DAY) and
        r.sender = s.sender
GROUP BY s.sender, Days.n
ORDER BY s.sender, Days.n DESC;

这使用returns 表来获取适当的发件人。如果您有另一个表，则可以使用它。

【讨论】：

感谢@Gordon Linoff，。完全符合我的要求