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];

对于SQL Server 2008 +使用MERGE而不是专有的更新…FROM语法有一些吸引力。

它不仅是标准SQL，因此更易于移植，而且在源端有多个连接行的情况下(因此在更新中可能使用多个不同的值)也会引发错误，而不是最终结果不确定。

MERGE INTO Sales_Import
   USING RetrieveAccountNumber
      ON Sales_Import.LeadID = RetrieveAccountNumber.LeadID
WHEN MATCHED THEN
   UPDATE 
      SET AccountNumber = RetrieveAccountNumber.AccountNumber;

然而，不幸的是，选择使用哪一种可能并不完全取决于首选的风格。SQL Server中MERGE的实现一直受到各种错误的困扰。Aaron Bertrand在这里整理了一份报告的清单。

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;

MS Sql

UPDATE  c4 SET Price=cp.Price*p.FactorRate FROM TableNamea_A c4
inner join TableNamea_B p on c4.Calcid=p.calcid 
inner join TableNamea_A cp on c4.Calcid=cp.calcid 
WHERE c4..Name='MyName';

Oracle 11 g

        MERGE INTO  TableNamea_A u 
        using
        (
                SELECT c4.TableName_A_ID,(cp.Price*p.FactorRate) as CalcTot 
                FROM TableNamea_A c4
                inner join TableNamea_B p on c4.Calcid=p.calcid 
                inner join TableNamea_A cp on c4.Calcid=cp.calcid 
                WHERE p.Name='MyName' 
        )  rt
        on (u.TableNamea_A_ID=rt.TableNamea_B_ID)
        WHEN MATCHED THEN
        Update set Price=CalcTot  ;

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
)

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)