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

Oracle 11 g

merge into Sales_Import
using RetrieveAccountNumber
on (Sales_Import.LeadId = RetrieveAccountNumber.LeadId)
when matched then update set Sales_Import.AccountNumber = RetrieveAccountNumber.AccountNumber;

2019-01-21 16:40:30

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

这是Mysql和Maria DB最简单和最好的

UPDATE table2, table1 SET table2.by_department = table1.department WHERE table1.id = table2.by_id

注意:如果您遇到以下错误基于您的Mysql/Maria DB版本“错误代码:1175。您正在使用安全更新模式，并且您试图更新一个没有使用KEY列的WHERE的表。要禁用安全模式，请切换首选项中的选项。

然后像这样运行代码

SET SQL_SAFE_UPDATES=0;
UPDATE table2, table1 SET table2.by_department = table1.department WHERE table1.id = table2.by_id

2021-03-08 11:55:03

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

MYSQL(这是我最喜欢的恢复所有特定列reasonId值的方法，基于主键id等价)

UPDATE `site` AS destination  
INNER JOIN `site_copy` AS backupOnTuesday 
      ON backupOnTuesday.`id` = destination.`id`
SET destdestination.`reasonId` = backupOnTuesday.`reasonId`

2021-01-06 09:46:14

下面是在SQL Server中对我有效的方法:

UPDATE [AspNetUsers] SET

[AspNetUsers].[OrganizationId] = [UserProfile].[OrganizationId],
[AspNetUsers].[Name] = [UserProfile].[Name]

FROM [AspNetUsers], [UserProfile]
WHERE [AspNetUsers].[Id] = [UserProfile].[Id];

2018-09-07 12:49:12

基于ID匹配从一个表到另一个表的SQL更新

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