试试这个:

UPDATE
    Table_A
SET
    Table_A.AccountNumber = Table_B.AccountNumber ,
FROM
    dbo.Sales_Import AS Table_A
    INNER JOIN dbo.RetrieveAccountNumber AS Table_B
        ON Table_A.LeadID = Table_B.LeadID 
WHERE
    Table_A.LeadID = Table_B.LeadID

下面的SQL有人建议，不工作在SQL Server。这个语法让我想起了我以前的学校类:

UPDATE table2 
SET table2.col1 = table1.col1, 
table2.col2 = table1.col2,
...
FROM table1, table2 
WHERE table1.memberid = table2.memberid

不建议使用NOT IN或NOT EXISTS的所有其他查询。显示null是因为OP将整个数据集与较小的子集进行比较，当然会有匹配问题。这必须通过使用正确的JOIN编写正确的SQL来解决，而不是使用NOT IN来逃避问题。在这种情况下，使用NOT IN或NOT EXISTS可能会遇到其他问题。

我投票给最上面的，这是一种传统的方法，通过加入SQL Server来根据另一个表更新一个表。就像我说的，你不能在SQL Server的同一个UPDATE语句中使用两个表，除非你先连接它们。

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)  

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
);

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:

甲骨文

use

UPDATE suppliers
SET supplier_name = (SELECT customers.customer_name
                     FROM customers
                     WHERE customers.customer_id = suppliers.supplier_id)
WHERE EXISTS (SELECT customers.customer_name
              FROM customers
              WHERE customers.customer_id = suppliers.supplier_id);