Question

我有一张表格：

id | firstname | lastname | userid
---+-----------+------------------------
1  | john      | smith    | 545868-5434-343435-35353
2  | adam      | finger   | 545868-5434-343435-35353
3  | teri      | marti    | 545868-5434-343435-35353
4  | pei       | port     | 545868-5434-343435-35353

在数据库中我有很多userid我需要填充与firstname lastname完全相同的userid和Database >

这是我的SQl查询

SELECT  

cID, c.firstname,c.lastname,

[s].UserID,c.OwnerID

FROM  
Customer INNER JOIN [s] ON c.OwnerID = [s].UserID AND c.AssignedtoID =
[s].UserID AND c.CreatedByUserID = [s].UserID

AssignedtoID与UserID

相同

Answer 1

这对你有帮助。？

    Create table #tmpCustomer (id int, firstname VARCHAR(50),lastname VARCHAR(50),userid VARCHAR(100))


    INSERT INTO #tmpCustomer 
    SELECT 1, 'john','smith','545868-5434-343435-35353'
    union
    SELECT 2,'adam','finger','545868-5434-343435-35353'
    union
    SELECT 3,'teri','marti','545868-5434-343435-35353'
    union
    SELECT 4, 'pei','port','545868-5434-343435-35353'
    union
    SELECT 5, 'abc','xyz','545868-5434-343435-35354'
    union
    SELECT 6, 'mno','ert','545868-5434-343435-35354'


    --select * from #tmpCustomer

    ;with cte1 AS(Select row_number()over(partition by userid order by id) rn,* from #tmpCustomer ),
          cte2 AS (select * from cte1 where rn=1 ) 

    update t
    set t.firstname=c.firstname
    from #tmpCustomer t
    JOIN cte2 c on t.userid=c.userid 

    select * from #tmpCustomer

    drop table #tmpCustomer

Answer 2

我不知道我是否理解您的问题，请尝试下面的解决方案

DECLARE @cust as table（firstname varchar（20），lastname varchar（20））

插入@cust 值（＆＃39;苏珊＆＃39;＆＃39;史密斯＆＃39）

将@id声明为表（id int identity，任何varchar（20），row_inserted datetime2 default（cast（sysdatetime（）as datetime2）））

INSERT @id （任何，row_inserted）

SELECT＆＃39; x＆＃39; ＆＃39; 20180305＆＃39; 联合所有选择＆＃39; y＆＃39; 20180305＆＃39; 联合所有选择＆＃39;＆＃39; 20180305＆＃39;

选择s.id，c.firstname， c.lastname 来自@id as s 交叉加入@cust as c

如何将给定的SQL记录填充到所有UserID

2 个答案: