按日期范围内的日期汇总数据，结果集中没有日期间隔

小编典典

按日期范围内的日期汇总数据，结果集中没有日期间隔

sql

我有一个带有卖单的表格，我想列出COUNT两个日期之间每天的卖单数量，并且不留日期间隔。

这是我目前所拥有的：

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;

我得到的结果如下：

6     May 1
14    May 4
1     May 5
8     Jun 2
5     Jun 15

但是我想要得到的是：

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

那可能吗？

阅读 290

2021-03-10

共1个答案

小编典典

动态创建日期范围并针对您的订单表进行合并：-

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

这可以处理多达1000天的日期范围。

请注意，根据您用于日期的字段类型，可以轻松提高效率。

编辑-根据要求，以获取每月的订单数：-

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

2021-03-10