列值中获取第一个非空的值

标题是否符合网友的问题宗旨，另外讨论，暂且如此。想了妥解问题，还得看原讨论题。

这是一个网上的问题如下，

;with temp as 
(
    select '63738893' repair_no,'20190504' report_date,'HES2418819040700003'service_sheet_no,null sno,'467769309410' rno union all
    select '63738893' repair_no,'20190504' report_date,'HES2418819040700003'service_sheet_no,'467769309411' sno,null rno union all
    select '63738793' repair_no,'20190508' report_date,'HES2418819040700003'service_sheet_no,'467769309411' sno,null rno
)select * from temp
--以上原始数据
--以下想要的结果
;with temp as 
(
    select '63738893' repair_no,'20190504' report_date,'HES2418819040700003'service_sheet_no,'467769309411' sno,'467769309410' rno union all
    --select '63738893' repair_no,'20190504' report_date,'HES2418819040700003'service_sheet_no,'467769309411' sno,null rno union all
    select '63738793' repair_no,'20190508' report_date,'HES2418819040700003'service_sheet_no,'467769309411' sno,null rno
)select * from temp

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下面是Insus.NET实现方法。另创建一张临时表，比网友的数据表添加一个字段ID，删除一些与问题无关的字段。

在MS SQL Server 2017版本中实现。

Insus.NET的方法是使用 ROW_NUMBER 和 PARTITION 时行分组：

先分析一列[sno]，看看：

;WITH s AS
(
SELECT ROW_NUMBER() OVER(PARTITION BY [repair_no] ORDER BY [id],[sno]) AS [ROW_NUM],
    [id],
    [repair_no],
    [sno]
    FROM #T
    WHERE [sno] IS NOT NULL
)

SELECT [ROW_NUM],[id],[repair_no],[sno] FROM s;

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另一列[rno]:

;WITH
r AS
(
    SELECT ROW_NUMBER() OVER(PARTITION BY [repair_no] ORDER BY [id],[rno]) AS [ROW_NUM],
    [id],
    [repair_no],
    [rno]
    FROM #T
    WHERE [rno] IS NOT NULL
)
SELECT [ROW_NUM],[id],[repair_no],[rno] FROM r;

Source Code

以上加个ID列，主要是为了让大家看到它的排序，拿到的是第一列非空的值。网友的问题，直接按[repair_no]排序即可。

下面代码是把上面2列合并在一起。

;WITH s AS
(
SELECT ROW_NUMBER() OVER(PARTITION BY [repair_no] ORDER BY [id],[sno]) AS [ROW_NUM],
    [id],
    [repair_no],
    [sno]
    FROM #T
    WHERE [sno] IS NOT NULL
),
r AS
(
    SELECT ROW_NUMBER() OVER(PARTITION BY [repair_no] ORDER BY [id],[rno]) AS [ROW_NUM],
    [id],
    [repair_no],
    [rno]
    FROM #T
    WHERE [rno] IS NOT NULL
)
SELECT s.[repair_no],[sno],[rno] FROM s
INNER JOIN r  on (s.[repair_no] = r.[repair_no])
WHERE s.[ROW_NUM] = 1 AND r.ROW_NUM = 1;

Source Code

使用色彩来引示可以看到明白：

把以上方法去解决网友的问题，却得到另外一个结果：

对比一下，原来空值也应该有，就是当一个值都没有时，才用空值填充。

看来得改写一下程序，创建临时表，存储结果。

2个字段分别处理，把结果MERGE来合并至临时表中：

CREATE TABLE #ok_result([repair_no] INT,[sno] nvarchar(50),[rno] NVARCHAR(50))

;with temp as 
(
    select '63738893' repair_no,'20190504' report_date,'HES2418819040700003'service_sheet_no,null sno,'467769309410' rno union all
    select '63738893' repair_no,'20190504' report_date,'HES2418819040700003'service_sheet_no,'467769309411' sno,null rno union all
    select '63738793' repair_no,'20190508' report_date,'HES2418819040700003'service_sheet_no,'467769309411' sno,null rno
),s AS
(
SELECT ROW_NUMBER() OVER(PARTITION BY [repair_no] ORDER BY repair_no) AS [ROW_NUM],    
    [repair_no],
    [sno]
    FROM temp
    WHERE [sno] IS NOT NULL
)
 MERGE #ok_result AS Target
    USING (SELECT [repair_no],[sno] FROM s WHERE [ROW_NUM] = 1) AS Source
    ON (Target.[repair_no] = Source.[repair_no])
 
    WHEN MATCHED THEN
        UPDATE SET target.[sno] = source.[sno]
 
    WHEN NOT MATCHED BY TARGET THEN
        INSERT ([repair_no],[sno]) VALUES ([repair_no],[sno]);  

        
;with temp as 
(
    select '63738893' repair_no,'20190504' report_date,'HES2418819040700003'service_sheet_no,null sno,'467769309410' rno union all
    select '63738893' repair_no,'20190504' report_date,'HES2418819040700003'service_sheet_no,'467769309411' sno,null rno union all
    select '63738793' repair_no,'20190508' report_date,'HES2418819040700003'service_sheet_no,'467769309411' sno,null rno
),r AS
(
SELECT ROW_NUMBER() OVER(PARTITION BY [repair_no] ORDER BY repair_no) AS [ROW_NUM],    
    [repair_no],
    [rno]
    FROM temp
    WHERE [rno] IS NOT NULL
)
 MERGE #ok_result AS Target
    USING (SELECT [repair_no],[rno] FROM r WHERE [ROW_NUM] = 1) AS Source
    ON (Target.[repair_no] = Source.[repair_no])
 
    WHEN MATCHED THEN
        UPDATE SET target.[rno] = source.[rno]
 
    WHEN NOT MATCHED BY TARGET THEN
        INSERT ([repair_no],[sno]) VALUES ([repair_no],[rno]);  



SELECT [repair_no],[sno],[rno] FROM #ok_result

Source Code