我正在尝试将license_no字段更新为null 如果 license_no重复并且具有最早的termination_date 和，如果该termination_date不是空值。下面的SQL查询是我到目前为止所拥有的。如何确定update语句中重复项的min termination_date？
update driver
set license_no = NULL
where license_no IN (SELECT license_no
FROM driver
GROUP BY license_no
HAVING ( Count(license_no) > 1 ))
答案 0 :(得分：1)
使用window function。试试这个。
UPDATE A
SET license_no = NULL
FROM driver A
JOIN (SELECT Min(termination_date) termination_date,
license_no
FROM (SELECT Row_number()OVER(partition BY license_no ORDER BY termination_date desc) rn,
*
FROM driver) o
WHERE rn > 1
GROUP BY license_no) B
ON a.license_no = b.license_no
AND a.termination_date = b.termination_date
答案 1 :(得分：0)
我将使用表变量来获取一些示例数据。我将添加许可证编号和日期的各种组合以捕捉所有可能性。
此外，我将在您的架构中添加并使用ID列。我的解决方案我需要分别识别每一行。仅通过许可证号识别行是不够的。
DECLARE @TDrivers TABLE (ID int NOT NULL PRIMARY KEY, license_no varchar(50) NULL, termination_date date NULL);
insert into @TDrivers values (1, NULL, '20130101');
insert into @TDrivers values (2, NULL, '20130101');
insert into @TDrivers values (3, NULL, '20130102');
insert into @TDrivers values (4, NULL, '20130103');
insert into @TDrivers values (5, NULL, '20130104');
-- only one license with this number
insert into @TDrivers values (6, '01', '20140101');
-- only one license with this number
insert into @TDrivers values (7, '02', NULL);
-- two licenses with this number, different dates
insert into @TDrivers values (8, '03', '20140301');
insert into @TDrivers values (9, '03', '20140302');
-- two licenses with this number, same dates
insert into @TDrivers values (10, '04', '20140401');
insert into @TDrivers values (11, '04', '20140401');
-- two licenses with this number, dates are not null and null
insert into @TDrivers values (12, '05', '20140501');
insert into @TDrivers values (13, '05', NULL);
-- two licenses with this number, both dates are null
insert into @TDrivers values (14, '06', NULL);
insert into @TDrivers values (15, '06', NULL);
-- three licenses with this number, dates are unique and not null
insert into @TDrivers values (16, '07', '20140701');
insert into @TDrivers values (17, '07', '20140702');
insert into @TDrivers values (18, '07', '20140703');
-- three licenses with this number, dates are not unique and not null
insert into @TDrivers values (19, '08', '20140801');
insert into @TDrivers values (20, '08', '20140801');
insert into @TDrivers values (21, '08', '20140802');
-- dates are unique and have one null
insert into @TDrivers values (22, '09', '20140901');
insert into @TDrivers values (23, '09', '20140902');
insert into @TDrivers values (24, '09', '20140903');
insert into @TDrivers values (25, '09', NULL);
-- dates are unique and have several nulls
insert into @TDrivers values (26, '10', '20141001');
insert into @TDrivers values (27, '10', '20141002');
insert into @TDrivers values (28, '10', '20141003');
insert into @TDrivers values (29, '10', NULL);
insert into @TDrivers values (30, '10', NULL);
insert into @TDrivers values (31, '10', NULL);
-- dates are not unique and have one null
insert into @TDrivers values (32, '11', '20141101');
insert into @TDrivers values (33, '11', '20141101');
insert into @TDrivers values (34, '11', '20141102');
insert into @TDrivers values (35, '11', '20141103');
insert into @TDrivers values (36, '11', NULL);
-- dates are not unique and have several nulls
insert into @TDrivers values (37, '12', '20141201');
insert into @TDrivers values (38, '12', '20141201');
insert into @TDrivers values (39, '12', '20141202');
insert into @TDrivers values (40, '12', '20141203');
insert into @TDrivers values (41, '12', NULL);
insert into @TDrivers values (42, '12', NULL);
insert into @TDrivers values (43, '12', NULL);
SELECT * FROM @TDrivers ORDER BY ID;
首先，我们会找到多次出现的所有许可证号码：
WITH
CTE_DuplicatedLicenses
AS
(
SELECT license_no
FROM @TDrivers AS d
WHERE license_no IS NOT NULL
-- there is no need to update license numbers that are NULL already
GROUP BY license_no
HAVING (COUNT(license_no) > 1)
)
然后，对于每个重复的许可证号，我们正在寻找一个具有最小终止日期的行，而不是NULL。要在多个行具有相同的终止日期时解决歧义，请使用ID中的ORDER BY列。
,CTE_LicensesWithDates
AS
(
SELECT
CTE_Dates.ID
FROM
CTE_DuplicatedLicenses
CROSS APPLY
(
SELECT TOP (1) d.ID
FROM @TDrivers AS d
WHERE
d.license_no = CTE_DuplicatedLicenses.license_no
AND d.termination_date IS NOT NULL
ORDER BY
d.termination_date, d.ID
) AS CTE_Dates
)
最后一个CTE为我们提供了应更新的所有IDs：
update @TDrivers
set license_no = NULL
where ID IN
(
SELECT ID
FROM CTE_LicensesWithDates
);
SELECT * FROM @TDrivers ORDER BY ID;
这是原始表：
ID license_no termination_date
1 NULL 2013-01-01
2 NULL 2013-01-01
3 NULL 2013-01-02
4 NULL 2013-01-03
5 NULL 2013-01-04
6 01 2014-01-01
7 02 NULL
8 03 2014-03-01
9 03 2014-03-02
10 04 2014-04-01
11 04 2014-04-01
12 05 2014-05-01
13 05 NULL
14 06 NULL
15 06 NULL
16 07 2014-07-01
17 07 2014-07-02
18 07 2014-07-03
19 08 2014-08-01
20 08 2014-08-01
21 08 2014-08-02
22 09 2014-09-01
23 09 2014-09-02
24 09 2014-09-03
25 09 NULL
26 10 2014-10-01
27 10 2014-10-02
28 10 2014-10-03
29 10 NULL
30 10 NULL
31 10 NULL
32 11 2014-11-01
33 11 2014-11-01
34 11 2014-11-02
35 11 2014-11-03
36 11 NULL
37 12 2014-12-01
38 12 2014-12-01
39 12 2014-12-02
40 12 2014-12-03
41 12 NULL
42 12 NULL
43 12 NULL
这是更新的表格：
ID license_no termination_date
1 NULL 2013-01-01
2 NULL 2013-01-01
3 NULL 2013-01-02
4 NULL 2013-01-03
5 NULL 2013-01-04
6 01 2014-01-01
7 02 NULL
8 NULL 2014-03-01
9 03 2014-03-02
10 NULL 2014-04-01
11 04 2014-04-01
12 NULL 2014-05-01
13 05 NULL
14 06 NULL
15 06 NULL
16 NULL 2014-07-01
17 07 2014-07-02
18 07 2014-07-03
19 NULL 2014-08-01
20 08 2014-08-01
21 08 2014-08-02
22 NULL 2014-09-01
23 09 2014-09-02
24 09 2014-09-03
25 09 NULL
26 NULL 2014-10-01
27 10 2014-10-02
28 10 2014-10-03
29 10 NULL
30 10 NULL
31 10 NULL
32 NULL 2014-11-01
33 11 2014-11-01
34 11 2014-11-02
35 11 2014-11-03
36 11 NULL
37 NULL 2014-12-01
38 12 2014-12-01
39 12 2014-12-02
40 12 2014-12-03
41 12 NULL
42 12 NULL
43 12 NULL
要运行此示例，只需将所有代码块放在一个脚本中即可。
您可以在结果中看到ID为8,10,12的行已更新。
ID为14,15的行未更新，因为它们都具有NULL日期。
对于每个其他许可证编号，您可以看到只有一行设置为NULL，即使多行具有相同的日期。
答案 2 :(得分：0)
我会使用这样的标准SQL：
UPDATE
driver d
SET
license_no = null
WHERE
Exists ( SELECT
d2.license_no
FROM
driver d2
WHERE
d2.license_no = d.license_no
GROUP BY
d2.license_no
HAVING
Count(d2.license_no) > 1
) AND
Exists ( SELECT
d3.license_no
FROM
driver d3
WHERE
d3.license_no = d.license_no AND
d3.termination_date = d.termination_date AND
d3.termination_date IS NOT NULL AND
d3.termination_date IN ( SELECT
Min(d4.termination_date)
FROM
driver d4
WHERE
d4.license_no = d.license_no )
)
第一个Exists()函数检查重复项。
第二个Exists()函数检查termination_date约束。
希望它有所帮助！