基于ID匹配从一个表到另一个表的SQL更新

我有一个有帐号和卡号的数据库。我将这些匹配到一个文件，以将任何卡号更新为帐号，这样我只使用帐号。

我创建了一个将表链接到帐户/卡数据库的视图，以返回table ID和相关的帐号，现在我需要更新那些ID与account number匹配的记录。

这是Sales_Import表，其中的帐号字段需要更新:

LeadID	AccountNumber
147	5807811235
150	5807811326
185	7006100100007267039

这是RetrieveAccountNumber表，我需要从这里更新:

LeadID	AccountNumber
147	7006100100007266957
150	7006100100007267039

我尝试了下面的方法，但到目前为止运气都不佳:

UPDATE [Sales_Lead].[dbo].[Sales_Import] 
SET    [AccountNumber] = (SELECT RetrieveAccountNumber.AccountNumber 
                          FROM   RetrieveAccountNumber 
                          WHERE  [Sales_Lead].[dbo].[Sales_Import]. LeadID = 
                                                RetrieveAccountNumber.LeadID)

它将卡号更新为帐号，但是帐号被NULL替换

当前回答

在同一个表内更新:

  DECLARE @TB1 TABLE
    (
        No Int
        ,Name NVarchar(50)
        ,linkNo int
    )

    DECLARE @TB2 TABLE
    (
        No Int
        ,Name NVarchar(50)
        ,linkNo int
    )

    INSERT INTO @TB1 VALUES(1,'changed person data',  0);
    INSERT INTO @TB1 VALUES(2,'old linked data of person', 1);

INSERT INTO @TB2 SELECT * FROM @TB1 WHERE linkNo = 0


SELECT * FROM @TB1
SELECT * FROM @TB2


    UPDATE @TB1 
        SET Name = T2.Name
    FROM        @TB1 T1
    INNER JOIN  @TB2 T2 ON T2.No = T1.linkNo

    SELECT * FROM @TB1

2012-12-06 12:24:29

其他回答

对于Oracle SQL，请尝试使用别名

UPDATE Sales_Lead.dbo.Sales_Import SI 
SET SI.AccountNumber = (SELECT RAN.AccountNumber FROM RetrieveAccountNumber RAN WHERE RAN.LeadID = SI.LeadID);

2022-02-22 08:08:03

它与postgresql一起工作

UPDATE application
SET omts_received_date = (
    SELECT
        date_created
    FROM
        application_history
    WHERE
        application.id = application_history.application_id
    AND application_history.application_status_id = 8
);

2015-05-14 09:30:28

在同一个表内更新:

  DECLARE @TB1 TABLE
    (
        No Int
        ,Name NVarchar(50)
        ,linkNo int
    )

    DECLARE @TB2 TABLE
    (
        No Int
        ,Name NVarchar(50)
        ,linkNo int
    )

    INSERT INTO @TB1 VALUES(1,'changed person data',  0);
    INSERT INTO @TB1 VALUES(2,'old linked data of person', 1);

INSERT INTO @TB2 SELECT * FROM @TB1 WHERE linkNo = 0


SELECT * FROM @TB1
SELECT * FROM @TB2


    UPDATE @TB1 
        SET Name = T2.Name
    FROM        @TB1 T1
    INNER JOIN  @TB2 T2 ON T2.No = T1.linkNo

    SELECT * FROM @TB1

2012-12-06 12:24:29

为未来的开发人员提供的一般答案。

SQL Server

UPDATE 
     t1
SET 
     t1.column = t2.column
FROM 
     Table1 t1 
     INNER JOIN Table2 t2 
     ON t1.id = t2.id;

Oracle(和SQL Server)

UPDATE 
     t1
SET 
     t1.colmun = t2.column 
FROM 
     Table1 t1, 
     Table2 t2 
WHERE 
     t1.ID = t2.ID;

MySQL

UPDATE 
     Table1 t1, 
     Table2 t2
SET 
     t1.column = t2.column 
WHERE
     t1.ID = t2.ID;

2016-08-29 17:59:14

我对foo也有同样的问题。对于在bar中没有匹配键的foo行，New被设置为null。我在Oracle做了类似的事情:

update foo
set    foo.new = (select bar.new
                  from bar 
                  where foo.key = bar.key)
where exists (select 1
              from bar
              where foo.key = bar.key)

2009-04-29 18:32:22

基于ID匹配从一个表到另一个表的SQL更新

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