问题描述
我正在学习递归函数,因为我需要在几天的时间内每天提取一行.这是我当前的数据
I'm learning about recursive functions, Since I need to extract a row for each day in a range of days. This is my current data
+----+------------+------------+
| id | from | to |
+----+------------+------------+
| 1 | 09-20-2019 | 09-25-2019 |
+----+------------+------------+
目标是按如下方式接收我的数据
The goal is to receive my data as follows
+----+------------+
| id | date |
+----+------------+
| 1 | 09-20-2019 |
| 1 | 09-21-2019 |
| 1 | 09-22-2019 |
| 1 | 09-23-2019 |
| 1 | 09-24-2019 |
| 1 | 09-25-2019 |
+----+------------+
我正在跟踪此处显示的示例: https://stackoverflow.com/a/54538866/1731057 但是由于某种原因,我的递归函数正在寻找'cte'表.
I'm following an example seen here: https://stackoverflow.com/a/54538866/1731057But for some reason my recursive function is looking for the 'cte' table.
WITH cte AS (
SELECT date_from
FROM event_dates
UNION ALL
SELECT DATE_ADD(event_dates.date_from, INTERVAL 1 DAY)
FROM cte
WHERE DATE_ADD(event_dates.date_from, INTERVAL 1 DAY) <= event_dates.date_until
)
select * FROM cte;
推荐答案
递归CTE的结构已关闭,并且联合的上半部分应该是种子基础案例.然后,递归部分应将前一天的输入值增加一天:
The structure of your recursive CTE is off, and the upper half of the union should be a seed base case. Then, the recursive part should add one day to the previous incoming value:
WITH RECURSIVE cte (n, dt) AS (
SELECT 1, '2019-09-20'
UNION ALL
SELECT n + 1, TIMESTAMPADD(DAY, n, '2019-09-20') FROM cte WHERE n <= 5
)
SELECT * FROM cte;
值得注意的是,我们在这里使用TIMESTAMPADD()
来解决INTERVAL
表达式的问题,该表达式不能真正使用变量.
Of note, we use TIMESTAMPADD()
here to get around the problem of the INTERVAL
expression, which can't really take a variable.
如果要使用此方法来生成一系列与表中的from和to值匹配的日期,则可以尝试联接:
If you want to use this approach to generate a series of dates which matches the from and to values in your table, then you can try a join:
SELECT
t1.dt
FROM cte t1
INNER JOIN yourTable t2
ON t1.dt BETWEEN t2.from_date AND t2.to_date;
以这种方式使用时,递归CTE充当日历表.
When used this way, the recursive CTE is acting as a calendar table.
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