Sql 仅从每组中选择最上面的行

我对下表有问题。我只能选择具有最大USCO_DFR和USCO_AHO='1'的用户。因此，从下面的示例中可以看出：

+----------+-------------------------+----------+
| USER_KEY |        USCO_DFR         | USCO_AHO |
+----------+-------------------------+----------+
|        1 | 2018-06-01 00:00:00.000 | NULL     |
|        1 | 2018-03-05 00:00:00.000 | 1        |
|        1 | 2018-02-10 00:00:00.000 | NULL     |
|        2 | 2018-07-10 00:00:00.000 | 1        |
|        2 | 2018-04-05 00:00:00.000 | NULL     |
|        2 | 2018-01-15 00:00:00.000 | NULL     |
|        3 | 2018-09-10 00:00:00.000 | 1        |
|        3 | 2018-05-05 00:00:00.000 | NULL     |
|        3 | 2018-04-15 00:00:00.000 | NULL     |
+----------+-------------------------+----------+

只应选择用户密钥=2,3

预期产出：

+----------+-------------------------+----------+
| USER_KEY |        USCO_DFR         | USCO_AHO |
+----------+-------------------------+----------+
|        2 | 2018-07-10 00:00:00.000 |        1 |
|        3 | 2018-09-10 00:00:00.000 |        1 |
+----------+-------------------------+----------+

此查询对结果进行排序：

SELECT * FROM @BAUSCO ORDER BY USER_KEY, USCO_DFR DESC

但我不知道如何从结果中选择那些用户密钥？基本上，我必须从每个集合中只选择最上面的一行，这满足条件USCO_AHO='1'

DECLARE @BAUSCO TABLE
(
    USER_KEY INT,
    USCO_DFR DATETIME,
    USCO_AHO CHAR(1)
)

INSERT @BAUSCO(USER_KEY, USCO_DFR, USCO_AHO)
VALUES (1, '2018-02-10', NULL),
(1, '2018-03-05', '1'),
(1, '2018-06-01', NULL),

(2, '2018-01-15', NULL),
(2, '2018-04-05', NULL),
(2, '2018-07-10', '1'),

(3, '2018-04-15', NULL),
(3, '2018-05-05', NULL),
(3, '2018-09-10', '1')

---

我们可以在此处使用

行数

来针对每个用户的最大

USCO\u DFR

记录：

WITH cte AS (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY USER_KEY ORDER BY USCO_DFR DESC) rn
    FROM @BAUSCO
)

SELECT USER_KEY, USCO_DFR, USCO_AHO
FROM cte
WHERE USCO_AHO = 1 and rn = 1;

这将选择其最大

USCO\u DFR

值与

USCO\u AHO

值为1的同一记录一致的所有用户记录。

尝试此操作

SELECT A.* FROM @BAUSCO A INNER JOIN
(
     SELECT USER_KEY, Max(USCO_DFR) MaxUSCO_DFR
     FROM @BAUSCO
     GROUP BY USER_KEY
) B
ON A.USER_KEY = B.USER_KEY AND A.USCO_DFR = B.MaxUSCO_DFR
WHERE A.USCO_AHO = 1 

---

您可以将行号与CTE一起使用

;with cte as
(
select ROW_NUMBER()  over (partition by USER_KEY order by USCO_DFR desc) AS ROWNUM,*
from 
@BAUSCO
)
select USER_KEY, USCO_DFR, USCO_AHO from cte where ROWNUM=1 and USCO_AHO =1

---

以下查询应执行您想要的操作：

DECLARE @BAUSCO TABLE
(
    USER_KEY INT,
    USCO_DFR DATETIME,
    USCO_AHO CHAR(1)
)

INSERT @BAUSCO(USER_KEY, USCO_DFR, USCO_AHO)
VALUES (1, '2018-02-10', NULL),
(1, '2018-03-05', '1'),
(1, '2018-06-01', NULL),

(2, '2018-01-15', '1'),
(2, '2018-04-05', NULL),
(2, '2018-07-10', '1'),

(3, '2018-04-15', '1'),
(3, '2018-05-05', NULL),
(3, '2018-09-10', '1')

SELECT USER_KEY, USCO_DFR, USCO_AHO FROM (
SELECT USER_KEY
    ,USCO_DFR
    ,USCO_AHO
    ,ROW_NUMBER() OVER (PARTITION BY USER_KEY ORDER BY (SELECT 1)) AS RNO
FROM @BAUSCO ) A
WHERE A.USCO_AHO = A.RNO AND A.USCO_AHO = 1

结果如下：

USER_KEY    USCO_DFR                  USCO_AHO
2           2018-01-15 00:00:00.000   1
3           2018-04-15 00:00:00.000   1

---

在简单的SQL中，使用GROUP BY with sub查询，可以得到预期的结果：

SELECT Q.USER_KEY, Q.USCO_DFR, B.USCO_AHO
FROM (
    SELECT USER_KEY, MAX(USCO_DFR) AS USCO_DFR
    FROM @BAUSCO
    GROUP BY USER_KEY
) Q 
JOIN @BAUSCO B ON B.USER_KEY = Q.USER_KEY AND B.USCO_DFR = Q.USCO_DFR
WHERE B.USCO_AHO = '1'

---

也许有点过分了，可以用来分解分析函数，但它们真是太方便了

SELECT * FROM
(SELECT
    *,
    MAX(USCO_DFR) OVER (PARTITION BY USER_KEY) AS MAX_DFR
  FROM
    @BAUSCO
) T
WHERE
  T.USCO_AHO = '1' 
  AND T.USCO_DFR = T.MAX_DFR

结果:

| USER_KEY |             USCO_DFR | USCO_AHO |              MAX_DFR |
|----------|----------------------|----------|----------------------|
|        2 | 2018-07-10T00:00:00Z |        1 | 2018-07-10T00:00:00Z |
|        3 | 2018-09-10T00:00:00Z |        1 | 2018-09-10T00:00:00Z |

---

只应选择用户2,3，因为他们的USCO_AHO='1'具有最大的USCO_DFR。您的查询返回三个结果。请参阅我的预期输出，我已添加此内容。您编辑的查询返回空表。你试过了吗？我觉得你的提问和你的概念很棒。仅此而已：ORDER BY USCO_DFR DESC->所以它是按USCO_DFR排序的，最大的结果在顶部。我非常喜欢分析函数，可以清楚地表达您想要实现的内容，并让SQL引擎决定如何实现。在这种情况下，我认为使用row_number和order稍微模糊了意图——您可以直接返回max值。它确实提出了如何处理USCO_DFR值完全相同的行的问题。@IanMcGowan您的答案也遇到了同样的问题。如果两个或多个记录绑定到最大值，您将得到多个结果。您的查询不会返回任何结果。空表。@FrenkyB我已经包括了完整的代码和结果

DECLARE @BAUSCO TABLE
(
    USER_KEY INT,
    USCO_DFR DATETIME,
    USCO_AHO CHAR(1)
)

INSERT @BAUSCO(USER_KEY, USCO_DFR, USCO_AHO)
VALUES (1, '2018-02-10', NULL),
(1, '2018-03-05', '1'),
(1, '2018-06-01', NULL),

(2, '2018-01-15', NULL),
(2, '2018-04-05', NULL),
(2, '2018-07-10', '1'),

(3, '2018-04-15', NULL),
(3, '2018-05-05', NULL),
(3, '2018-09-10', '1')

select * from @BAUSCO a
where USCO_DFR=(select MAX(USCO_DFR) from @BAUSCO b where a.USER_KEY=b.USER_KEY )
    and USCO_AHO=1