Mysql 按日期范围内的日期聚合数据，结果集中没有日期间隔

我有一个销售订单表，我想列出两个日期之间每天销售订单的

计数

，不留日期间隔

这就是我目前拥有的：

SELECT COUNT(*) as Norders, DATE_FORMAT(date, "%M %e") as sdate 
FROM ORDERS 
WHERE date <= NOW() 
  AND date >= NOW() - INTERVAL 1 MONTH 
GROUP BY DAY(date) 
ORDER BY date ASC;

但我想得到的是：

6     May 1
0     May 2
0     May 3
14    May 4
1     May 5
0     May 6
0     May 7
0     May 8
.....
0     Jun 1
8     Jun 2
.....
5     Jun 15

这可能吗？

首先创建一个

选择coalesce（计数（O*），0）作为Norders，DATE_格式（C.DATE，“%M%e”）作为sdate
从日历C
C.date=O.date上的左连接订单O
其中O.date=NOW（）-间隔1个月
按天分组（日期）
按日期订购ASC；

动态创建日期范围，并根据订单表加入该范围：-

SELECT sub1.sdate, COUNT(ORDERS.id) as Norders
FROM
(
    SELECT DATE_FORMAT(DATE_SUB(NOW(), INTERVAL units.i + tens.i * 10 + hundreds.i * 100 DAY), "%M %e") as sdate 
    FROM (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)units
    CROSS JOIN (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)tens
    CROSS JOIN (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9)hundreds
    WHERE DATE_SUB(NOW(), INTERVAL units.i + tens.i * 10 + hundreds.i * 100 DAY) BETWEEN DATE_SUB(NOW(), INTERVAL 1 MONTH) AND NOW()
) sub1
LEFT OUTER JOIN ORDERS
ON sub1.sdate = DATE_FORMAT(ORDERS.date, "%M %e")
GROUP BY sub1.sdate

这可以处理高达1000天的日期范围

请注意，根据您用于日期的字段类型，可以轻松地提高效率

编辑-根据要求，获取每月订单数量：-

SELECT aMonth, COUNT(ORDERS.id) as Norders
FROM
(
    SELECT DATE_FORMAT(DATE_SUB(NOW(), INTERVAL months.i MONTH), "%Y%m") as sdate, DATE_FORMAT(DATE_SUB(NOW(), INTERVAL months.i MONTH), "%M") as aMonth 
    FROM (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9 UNION SELECT 10 UNION SELECT 11)months
    WHERE DATE_SUB(NOW(), INTERVAL months.i MONTH) BETWEEN DATE_SUB(NOW(), INTERVAL 12 MONTH) AND NOW()
) sub1
LEFT OUTER JOIN ORDERS
ON sub1.sdate = DATE_FORMAT(ORDERS.date, "%Y%m")
GROUP BY aMonth

---

您需要生成一个虚拟（或物理）表，其中包含范围内的每个日期

这可以通过使用序列表按如下方式完成

SELECT mintime + INTERVAL seq.seq DAY AS orderdate
  FROM (
        SELECT CURDATE() - INTERVAL 1 MONTH AS mintime,
               CURDATE() AS maxtime
          FROM obs
       ) AS minmax
  JOIN seq_0_to_999999 AS seq ON seq.seq < TIMESTAMPDIFF(DAY,mintime,maxtime)

你可以在网站上找到关于这一切的详细解释

---

如果您使用的是MariaDB版本10+，这些序列表是内置的。

您有包含所有日期的日历表吗？@GordonLinoff否，我只有ORDERs表，其中只包含创建订单日期的行。@oliakaoil是的，完全正确。这就是重点。

ORDERS

表格不包含每天的行，而只包含创建订单的日期。这需要有一个

日历

表格，该表格在每年的每个月都有每天的行，对吗？@alexandernst是的，请按照答案中的链接进行操作，效率不高，IMHO。如果没有其他办法，我会把它作为B计划。@alexandernst我建议你对理货/数字/序列表的使用进行一些研究。RegardsNice技巧：）只有一个错误，它返回

1

而不是

0

我没有任何销售订单的日期。不应该这样做，但它确实需要COUNT（…）计算orders表（我假设id）中的一列，当没有订单时，该列的值将为NULL（COUNT和列名将忽略NULL值）。如果您使用COUNT（*），它会将没有订单的行计数为1。啊，的确如此。你说得对。我只是写了*而不是

id

。谢谢你，这很好用！就同一个话题再问一个问题。如果我想显示过去12个月每月订单的

计数

，我应该如何修改

sub1

子选项？增加了一个建议。+1指出MariaDB 10+内置了序列表。

SELECT aMonth, COUNT(ORDERS.id) as Norders
FROM
(
    SELECT DATE_FORMAT(DATE_SUB(NOW(), INTERVAL months.i MONTH), "%Y%m") as sdate, DATE_FORMAT(DATE_SUB(NOW(), INTERVAL months.i MONTH), "%M") as aMonth 
    FROM (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9 UNION SELECT 10 UNION SELECT 11)months
    WHERE DATE_SUB(NOW(), INTERVAL months.i MONTH) BETWEEN DATE_SUB(NOW(), INTERVAL 12 MONTH) AND NOW()
) sub1
LEFT OUTER JOIN ORDERS
ON sub1.sdate = DATE_FORMAT(ORDERS.date, "%Y%m")
GROUP BY aMonth

SELECT mintime + INTERVAL seq.seq DAY AS orderdate
  FROM (
        SELECT CURDATE() - INTERVAL 1 MONTH AS mintime,
               CURDATE() AS maxtime
          FROM obs
       ) AS minmax
  JOIN seq_0_to_999999 AS seq ON seq.seq < TIMESTAMPDIFF(DAY,mintime,maxtime)

SELECT IFNULL(orders.Norders,0) AS Norders, /* show zero instead of null*/
       DATE_FORMAT(alldates.orderdate, "%M %e") as sdate 
  FROM (
        SELECT mintime + INTERVAL seq.seq DAY AS orderdate
          FROM (
                SELECT CURDATE() - INTERVAL 1 MONTH AS mintime,
                       CURDATE() AS maxtime
                  FROM obs
               ) AS minmax
          JOIN seq_0_to_999999 AS seq 
                        ON seq.seq < TIMESTAMPDIFF(DAY,mintime,maxtime)
       ) AS alldates
  LEFT JOIN (
    SELECT COUNT(*) as Norders, DATE(date) AS orderdate
      FROM ORDERS 
    WHERE date <= NOW() 
      AND date >= NOW() - INTERVAL 1 MONTH 
    GROUP BY DAY(date) 
       ) AS orders ON alldates.orderdate = orders.orderdate
ORDER BY alldates.orderdate ASC

DROP TABLE IF EXISTS seq_0_to_9;
CREATE TABLE seq_0_to_9 AS
   SELECT 0 AS seq UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4
    UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9;

DROP VIEW IF EXISTS seq_0_to_999;
CREATE VIEW seq_0_to_999 AS (
SELECT (a.seq + 10 * (b.seq + 10 * c.seq)) AS seq
  FROM seq_0_to_9 a
  JOIN seq_0_to_9 b
  JOIN seq_0_to_9 c
);

DROP VIEW IF EXISTS seq_0_to_999999;
CREATE VIEW seq_0_to_999999 AS (
SELECT (a.seq + (1000 * b.seq)) AS seq
  FROM seq_0_to_999 a
  JOIN seq_0_to_999 b
);