如何在MySQL中按日期范围分组，而不跳过包含零项的日期

嗯，这可能很容易，但我似乎找不到任何解决办法

我有一个工作问题

SELECT 
    count(*), TIMESTAMPDIFF(DAY, buydate, NOW()) daydif
FROM
    order_item
WHERE
    buydate > NOW() - INTERVAL 4 DAY
GROUP BY daydif
ORDER BY daydif ASC

但结果是跳过了零项的日子

+----------+--------+
| count(*) | daydif |
+==========+========+
| 5        | 0      |
+----------+--------+
| 9        | 1      |
+----------+--------+
| 2        | 3      |
+----------+--------+

我想得到这个结果

+----------+--------+
| count(*) | daydif |
+==========+========+
| 5        | 0      |
+----------+--------+
| 9        | 1      |
+----------+--------+
| 0        | 2      |  //I want this row in the returned results
+----------+--------+
| 2        | 3      |
+----------+--------+

更新：

---

从答案和进一步的搜索来看，我似乎被迫创建一个helper表并加入其中。令人沮丧的是，我找不到像评论中提到的MariaDB那样的序列生成函数。是否有更优雅的方式的提示？

可能您没有任何物品，差异为2天。SQL无法创建某些东西，因为它不存在

您可以做的是使一个表具有一个由从0到最大的数字列表组成的长字段（在本例中为4），我们称之为tbl_diff，该字段为diff。在现有查询上左键连接tbl_diff以填补空白。请注意，我还将count（）更改为仅对设置了buydate的记录进行计数

SELECT 
    count(buydate), TIMESTAMPDIFF(DAY, buydate, NOW()) daydif
FROM
    tbl_diff
LEFT JOIN
    order_item on tbl_diff.diff=TIMESTAMPDIFF(DAY, buydate, NOW())
WHERE
    diff<=4
GROUP BY daydif
ORDER BY daydif ASC

---

正如Paul Spiegel所建议的那样，我最终使用了一些帮助表， 为了将来的参考和更好的帮助，这里是我用来生成表的代码

create table `stat_helper_calendar` (
  `date` DATE NOT NULL,
  PRIMARY KEY (`date`)
)
  select date_add('2015-01-01', INTERVAL  t3.c*1000 + t2.c*100 + t1.c*10 + t0.c DAY) as `date`
  from
    (select 0 c 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) t0,
    (select 0 c 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) t1,
    (select 0 c 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) t2,
    (select 0 c 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) t3

这个是整数

CREATE TABLE `stat_helper_num` (
  `num` INT NOT NULL,
  PRIMARY KEY (`num`)
) SELECT (t4.c * 10000 + t3.c * 1000 + t2.c * 100 + t1.c * 10 + t0.c) AS `num` FROM
  (SELECT 0 c 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) t0,
  (SELECT 0 c 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) t1,
  (SELECT 0 c 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) t2,
  (SELECT 0 c 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) t3,
  (SELECT 0 c 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) t4

---

您可以创建一个日历表，并使用日期将其与您的

order\u项目

表左键联接。您将了解如何创建日历表。在这种情况下，不需要日历表，只要有5个数字就足够了。在这种情况下，我同意。@PaulSpiegel，哇，这真是太聪明了！在较短范围内动态创建该表不是更好的选择吗？顺便说一句：如果您使用MariaDB，您可能想看看。刚刚测试过

选择seq as daydif从seq_0_到步骤_5_1

-效果很好。

buydate>NOW（）-间隔4天

应该在ON子句（IMHO）中@PaulSpiegel我不确定这是否会有任何区别。实际上，对于联接条件，它是一种冗余，因为如果差值大于4天，则联接将删除记录。如果daydif为2，则将选中

NULL>NOW（）-INTERVAL 4 DAY

，然后过滤掉该行。这就是行动计划的问题。《明镜周刊》你说得对。然而，我完全改变了where子句。在第二种情况下，甚至不需要where子句。

CREATE TABLE `stat_helper_num` (
  `num` INT NOT NULL,
  PRIMARY KEY (`num`)
) SELECT (t4.c * 10000 + t3.c * 1000 + t2.c * 100 + t1.c * 10 + t0.c) AS `num` FROM
  (SELECT 0 c 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) t0,
  (SELECT 0 c 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) t1,
  (SELECT 0 c 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) t2,
  (SELECT 0 c 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) t3,
  (SELECT 0 c 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) t4