Question

请考虑以下形状的表(wallet, client, date)，其中wallet和client是strings，而date是直到今天的任何日期。假设有人想获得唯一客户和客户今年的每个月的总和，按钱包和月份分组。我已经尝试了几种方法，包括分区，但是我无法使其正常工作。

结果将类似于：

+------------+---------+------------+---------+
|    Wallet  |  Unique |  Count     | Month   |
+------------+---------+------------+---------+
| Wallet1    | 2       |   5        |   1     | - For < 01022019
+------------+---------+------------+---------+
| Wallet1    | 3       |   7        |   2     | - For < 01032019
+------------+---------+------------+---------+

以此类推。我尝试了很多事情，甚至进行了分区，但都失败了。我只设法在日期以下使用WHERE使其变成几个小查询，这绝对糟糕，例如。

INSERT INTO 
    tmp_tbl_filtered
SELECT
    [Wallet],
    COUNT(DISTINCT [Client]) AS [unique],
    COUNT([Client]) AS [count],
    1 AS [month]
FROM
    tmp_tbl
WHERE [Date] < '20190201'
GROUP BY  [Wallet]
ORDER BY [Wallet]

INSERT INTO 
    tmp_tbl_filtered
SELECT
    [Wallet],
    COUNT(DISTINCT [Client]) AS [unique],
    COUNT([Client]) AS [count],
    2 AS [month]
FROM
    tmp_tbl
WHERE [Date] < '20190301'
GROUP BY  [Wallet]
ORDER BY [Wallet]

Answer 1

如果我做对了

SELECT [Wallet], [month]
   , SUM([unique]) OVER(PARTITION BY [Wallet] ORDER BY [month]) [unique_rt]
   , SUM([count]) OVER(PARTITION BY [Wallet] ORDER BY [month]) [count_rt]
FROM (
    --  group data of the current year 2019 by wallet and month
    SELECT 
        [Wallet],
        DATEPART(month,[Date]) AS [month],
        COUNT(DISTINCT [Client]) AS [unique],
        COUNT([Client]) AS [count]
    FROM
        tmp_tbl
    WHERE [Date] >= '20190101'
    GROUP BY [Wallet], DATEPART(month,[Date])
) t
ORDER BY [Wallet], [month]

Answer 2

我看不到累积部分在哪里发挥作用。

SELECT YEAR(date), MONTH(date), Wallet,
       COUNT(DISTINCT [Client]) AS [unique],
       COUNT([Client]) AS [count],
FROM tmp_tbl
WHERE [Date] < '20190301'
GROUP BY YEAR(date), MONTH(date), [Wallet]
ORDER BY Wallet, YEAR(date), MONTH(date)

T-SQL运行总和

2 个答案: