我们有一个表格,其中包含按日期计算的累计金额值,例如:
| date. | |
| 2015-03-01 | 1.23 |
| 2015-03-04 | 1.98 |
| 2015-03-06 | 2.23 |
And we have a table with all dates we need, for instance:
| date. |
| 2015-03-01 |
| 2015-03-02 |
| 2015-03-03 |
| 2015-03-04 |
| 2015-03-05 |
| 2015-03-06 |
We would like to get:
| date. | |
| 2015-03-01 | 1.23 |
| 2015-03-02 | 1.23 |
| 2015-03-03 | 1.23 |
| 2015-03-04 | 1.98 |
| 2015-03-05 | 1.98 |
| 2015-03-06 | 2.23 |
这可能只是使用SQL本身吗?
我正在努力想办法解决这个问题。
我尝试了一些交叉连接操作,但没有成功。
由于
答案 0 :(得分:4)
SELECT
[date],
amount
FROM (
SELECT
[date],
amount,
ROW_NUMBER() OVER(PARTITION BY [date] ORDER BY delta) AS win
FROM (
SELECT
dates_table.date AS [date],
DATEDIFF(TIMESTAMP(dates_table.date), TIMESTAMP(amounts_table.date)) AS delta,
amount
FROM amounts_table CROSS JOIN dates_table
WHERE DATEDIFF(TIMESTAMP(dates_table.date), TIMESTAMP(amounts_table.date)) >= 0
)
)
WHERE win = 1
ORDER BY 1
答案 1 :(得分:2)
我意识到 - 在这种特殊情况下,它应该更简单(更简单) 请尝试以下
SELECT
dates_table.date,
MAX(amounts.amount) OVER(ORDER BY dates_table.date) AS amount
FROM dates_table LEFT JOIN amounts_table
ON dates_table.date = amounts_table.date
你们中的一些人在采纳这个答案并思考它时遇到了问题 不起作用 - 我刚刚将以下示例添加到"证明"以上是有效的 :O)
SELECT
dates.date,
MAX(amounts.amount) OVER(ORDER BY dates.date) AS amount
FROM (
SELECT * FROM
(SELECT '2015-03-01' AS [date]),
(SELECT '2015-03-02' AS [date]),
(SELECT '2015-03-03' AS [date]),
(SELECT '2015-03-04' AS [date]),
(SELECT '2015-03-05' AS [date]),
(SELECT '2015-03-06' AS [date])
) AS dates
LEFT JOIN (
SELECT * FROM
(SELECT '2015-03-01' AS [date], 1.23 AS amount),
(SELECT '2015-03-04' AS [date], 1.98 AS amount),
(SELECT '2015-03-06' AS [date], 2.23 AS amount)
) AS amounts
ON dates.date = amounts.date
结果是:
Row dates_date amount
1 2015-03-01 1.23
2 2015-03-02 1.23
3 2015-03-03 1.23
4 2015-03-04 1.98
5 2015-03-05 1.98
6 2015-03-06 2.23