我想计算 2 个日期之间的所有条目(包括上周到今天),如果没有,请选择 0。目前它打印如下:
+-------+------------+
| items | SellDate |
+-------+------------+
| 1 | 2017-01-01 |
+-------+------------+
| 3 | 2017-01-02 |
+-------+------------+
| 1 | 2017-01-03 |
+-------+------------+
| 5 | 2017-01-06 |
+-------+------------+
但是,我需要这样打印的东西:
+-------+------------+
| items | SellDate |
+-------+------------+
| 1 | 2017-01-01 |
+-------+------------+
| 3 | 2017-01-02 |
+-------+------------+
| 1 | 2017-01-03 |
+-------+------------+
| 0 | 2017-01-04 |
+-------+------------+
| 0 | 2017-01-05 |
+-------+------------+
| 5 | 2017-01-06 |
+-------+------------+
| 0 | 2017-01-07 |
+-------+------------+
我的查询如下所示:
SELECT
COUNT(Item.id) AS Items,
DATE(Item.sold_at) AS SellDate
FROM Item
WHERE Item.sold_at IS NOT NULL AND Item.sold_at BETWEEN DATE(DATETIME('now', 'localtime', '-6 days')) AND DATE(DATETIME('now', 'localtime', '+1 day'))
GROUP BY SellDate
我做错了什么?
最佳答案
据我所知,如果没有 recursive common table expression,这是不可能的SQLite 3.8.3 及更高版本支持。使用相应的版本,您可以通过将日期范围与项目列表连接来实现:
WITH RECURSIVE dates(date) AS (
VALUES(DATE(DATETIME('now', 'localtime', '-6 days')))
UNION ALL
SELECT date(date, '+1 day')
FROM dates
WHERE date < DATE(DATETIME('now', 'localtime', '+1 day'))
)
SELECT
date,
COUNT(Item.id) AS Items
FROM
dates
LEFT JOIN
Item
ON
dates.date = Item.SellDate
GROUP BY
SellDate
关于sql - 选择日期范围之间的计数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46146972/