Question

SELECT COUNT(*),Date(createDate) FROM EEC_Order WHERE createDate
 BETWEEN DATE_SUB(CURDATE(),INTERVAL 7 DAY ) AND CURDATE()
    group by Date(createDate) ; 

COUNT(*)           Date(createDate)
  3                  2013-09-08
  1                  2013-09-11
  2                  2013-09-12

以上查询是从表中获取的，表示从当前日期开始的前7天（假设2013-09-13），基于创建的组。结果显示3行。在获取时间的同时，我想在上面的RESULT中添加日期和计数（不在数据库中）虚拟日期，这些日期并不像下面那样。请帮忙......如何写这个查询...

COUNT(*)    Date(createDate)
  0                  2013-09-06
  0                  2013-09-07
  3                  2013-09-08
  0                  2013-09-09
  0                  2013-09-10
  1                  2013-09-11
  2                  2013-09-12

问候
sakir esquareinfo

Answer 1

您需要一个日期列表和一个左外连接。以下是制定查询的一种方法：

SELECT COUNT(o.createDate), dates.thedate
FROM (select DATE_SUB(CURDATE(), INTERVAL 7 DAY ) as thedate union all
      select DATE_SUB(CURDATE(), INTERVAL 6 DAY ) as thedate union all
      select DATE_SUB(CURDATE(), INTERVAL 5 DAY ) as thedate union all
      select DATE_SUB(CURDATE(), INTERVAL 4 DAY ) as thedate union all
      select DATE_SUB(CURDATE(), INTERVAL 3 DAY ) as thedate union all
      select DATE_SUB(CURDATE(), INTERVAL 2 DAY ) as thedate union all
      select DATE_SUB(CURDATE(), INTERVAL 1 DAY ) as thedate union all
      select DATE_SUB(CURDATE(), INTERVAL 0 DAY ) as thedate
     ) dates left outer join
     EEC_Order o
     on Date(createDate) = dates.thedate
group by dates.thedate ;

编辑：

在MySQL中动态执行此操作并不容易，除非您在某处有numbers或dates表。以下采用这种方法，首先创建一个包含1000个数字的列表（带有where子句），然后进行连接和聚合：

select DATE_SUB(CURDATE(), INTERVAL n.n DAY ), count(e.CreateDate)
from (select (n1*100 + n2*10 + n3) as n
      from (select 0 as n 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
           ) n1 cross join
           (select 0 as n 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
           ) n2 cross join
           (select 0 as n 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
           ) n3
      where (n1*100 + n2*10 + n3) <= 30
     ) n left outer join
     EEC_Order o
     on Date(createDate) = DATE_SUB(CURDATE(), INTERVAL n.n DAY )
group by DATE_SUB(CURDATE(), INTERVAL n.n DAY );

在获取结果中附加虚拟数据

1 个答案: