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

如果以上答案不适合你，试试这个

Update Sales_Import A left join RetrieveAccountNumber B on A.LeadID = B.LeadID
Set A.AccountNumber = B.AccountNumber
where A.LeadID = B.LeadID 

似乎你正在使用MSSQL，然后，如果我没记错，它是这样做的:

UPDATE [Sales_Lead].[dbo].[Sales_Import] SET [AccountNumber] = 
RetrieveAccountNumber.AccountNumber 
FROM RetrieveAccountNumber 
WHERE [Sales_Lead].[dbo].[Sales_Import].LeadID = RetrieveAccountNumber.LeadID

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;

我想再补充一点。

不要用相同的值更新一个值，这会产生额外的日志记录和不必要的开销。 参见下面的例子-尽管链接了3条记录，但它只会在2条记录上执行更新。

DROP TABLE #TMP1
DROP TABLE #TMP2
CREATE TABLE #TMP1(LeadID Int,AccountNumber NVarchar(50))
CREATE TABLE #TMP2(LeadID Int,AccountNumber NVarchar(50))

INSERT INTO #TMP1 VALUES
(147,'5807811235')
,(150,'5807811326')
,(185,'7006100100007267039');

INSERT INTO #TMP2 VALUES
(147,'7006100100007266957')
,(150,'7006100100007267039')
,(185,'7006100100007267039');

UPDATE A
SET A.AccountNumber = B.AccountNumber
FROM
    #TMP1 A 
        INNER JOIN #TMP2 B
        ON
        A.LeadID = B.LeadID
WHERE
    A.AccountNumber <> B.AccountNumber  --DON'T OVERWRITE A VALUE WITH THE SAME VALUE

SELECT * FROM #TMP1

使用下面的查询块根据ID更新Table1和Table2:

UPDATE Sales_Import, RetrieveAccountNumber 
SET Sales_Import.AccountNumber = RetrieveAccountNumber.AccountNumber 
where Sales_Import.LeadID = RetrieveAccountNumber.LeadID;

这是解决这个问题最简单的方法。