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

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

其他回答

总结其他答案，关于如何仅在“匹配存在”时使用来自另一个表的数据更新目标表，有4种变体

查询和子查询:

update si
set    si.AccountNumber = (
    select ran.AccountNumber 
    from   RetrieveAccountNumber ran
    where  si.LeadID = ran.LeadID
)
from Sales_Import si
where exists (select * from RetrieveAccountNumber ran where ran.LeadID = si.LeadID)

内连接:

update si
set si.AccountNumber = ran.AccountNumber
from Sales_Import si inner join RetrieveAccountNumber ran on si.LeadID = ran.LeadID

交叉连接:

update si
set si.AccountNumber = ran.AccountNumber
from Sales_Import si, RetrieveAccountNumber ran
where si.LeadID = ran.LeadID

走：

merge into Sales_Import si
using RetrieveAccountNumber ran on si.LeadID = ran.LeadID 
when matched then update set si.accountnumber = ran.accountnumber;

所有的变体都是更少的琐碎和可以理解的，我个人更喜欢“内部连接”选项。但其中任何一种都可以使用，开发者必须根据自己的需求选择“更好的选项”

从性能的角度来看，使用join的变体更可取:

2022-12-22 11:40:16

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

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

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

PostgreSQL的:

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

2014-07-16 19:55:13

基于ID匹配从一个表到另一个表的SQL更新

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