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

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

其他回答

这将允许您根据未在另一个表中找到的列值更新表。

UPDATE table1 SET table1.column = 'some_new_val' WHERE table1.id IN (
        SELECT * 
        FROM (
                SELECT table1.id
                FROM  table1 
                LEFT JOIN table2 ON ( table2.column = table1.column ) 
                WHERE table1.column = 'some_expected_val'
                AND table12.column IS NULL
        ) AS Xalias
)

这将根据在两个表中找到的列值更新一个表。

UPDATE table1 SET table1.column = 'some_new_val' WHERE table1.id IN (
        SELECT * 
        FROM (
                SELECT table1.id
                FROM  table1 
                JOIN table2 ON ( table2.column = table1.column ) 
                WHERE table1.column = 'some_expected_val'
        ) AS Xalias
)

2015-04-02 15:27:50

根据匹配的id从一个表更新到另一个表

UPDATE 
     TABLE1 t1, 
     TABLE2 t2
SET 
     t1.column_name = t2.column_name 
WHERE
     t1.id = t2.id;

2020-12-21 13:25:30

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

谢谢你的回复。但我找到了一个解决办法。

UPDATE Sales_Import 
SET    AccountNumber = (SELECT RetrieveAccountNumber.AccountNumber 
                          FROM   RetrieveAccountNumber 
                          WHERE  Sales_Import.leadid =RetrieveAccountNumber.LeadID) 
WHERE Sales_Import.leadid = (SELECT  RetrieveAccountNumber.LeadID 
                             FROM   RetrieveAccountNumber 
                             WHERE  Sales_Import.leadid = RetrieveAccountNumber.LeadID)

2008-10-22 08:07:41

我相信一个带JOIN的UPDATE FROM会有帮助:

MS SQL

UPDATE
    Sales_Import
SET
    Sales_Import.AccountNumber = RAN.AccountNumber
FROM
    Sales_Import SI
INNER JOIN
    RetrieveAccountNumber RAN
ON 
    SI.LeadID = RAN.LeadID;

MySQL和MariaDB

UPDATE
    Sales_Import SI,
    RetrieveAccountNumber RAN
SET
    SI.AccountNumber = RAN.AccountNumber
WHERE
    SI.LeadID = RAN.LeadID;

2008-10-22 07:19:54

基于ID匹配从一个表到另一个表的SQL更新

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