使用来自SQL Server中另一个表的值更新表

我的数据库中有2个表。

表格1

-------------------------------------------------------------------------
| name | family | phone | email | gender | phone2 | address | birthdate |
-------------------------------------------------------------------------

表＃2

-----------------------------------------
| gender | address | phone | birthdate |
-----------------------------------------

在表＃1的列地址和PHONE2是空的和列性别和生日的值是相同的表＃2。

我怎样才能读取表＃2和更新数据的地址和PHONE2表＃1的值从表＃2 的地址和电话列时，性别和出生日期是各行中的一样吗？

例如：这是表＃1中的一些数据

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| name | family | phone | email | gender | phone2 | address | birthdate |
-------------------------------------------------------------------------
| john | doe    | 12345| t@t.com| Male  |         |         | 1980-01-01|
-------------------------------------------------------------------------
| mike | clark  | 65432| x@y.com| Male  |         |         | 1990-01-01|
-------------------------------------------------------------------------
| Sara | King   | 875465| a@b.com|Female|         |         | 1970-01-01|
-------------------------------------------------------------------------

这是表2中的一些数据

-----------------------------------------
| gender | address | phone | birthdate  |
-----------------------------------------
| Male   | 1704test|0457852|1980-01-01  |
-----------------------------------------
| Female | 1705abcs|0986532|1970-01-01  |
-----------------------------------------
| Male   | 1602cyzd|0326589|1990-01-01  |
-----------------------------------------

我想使用表＃2中的数据更新表＃1，并检查性别和生日，并使表＃1像

-------------------------------------------------------------------------
| name | family | phone | email | gender | phone2 | address | birthdate |
-------------------------------------------------------------------------
| john | doe    | 12345| t@t.com| Male   |0457852 |1704test | 1980-01-01|
-------------------------------------------------------------------------
| mike | clark  | 65432| x@y.com| Male   |0326589  |1602cyzd| 1990-01-01|
-------------------------------------------------------------------------
| Sara | King   | 875465| a@b.com|Female |0986532  |1705abcs| 1970-01-01|
-------------------------------------------------------------------------

我怎样才能做到这一点？

有很多方法可以达到您想要的结果。

不确定方法

（如果表2中的许多行与表1中的一行匹配）

UPDATE T1
SET    address = T2.address,
       phone2 = T2.phone
FROM   #Table1 T1
       JOIN #Table2 T2
         ON T1.gender = T2.gender
            AND T1.birthdate = T2.birthdate

或更简洁的形式

UPDATE #Table1
SET    address = #Table2.address,
       phone2 = #Table2.phone
FROM   #Table2
WHERE  #Table2.gender = #Table1.gender
       AND #Table2.birthdate = #Table1.birthdate 

或与CTE

WITH CTE
     AS (SELECT T1.address AS tgt_address,
                T1.phone2  AS tgt_phone,
                T2.address AS source_address,
                T2.phone   AS source_phone
         FROM   #Table1 T1
                INNER JOIN #Table2 T2
                  ON T1.gender = T2.gender
                     AND T1.birthdate = T2.birthdate)
UPDATE CTE
SET    tgt_address = source_address,
       tgt_phone = source_phone 

确定性方法

MERGE 会引发错误而不是接受不确定的结果

MERGE #Table1 T1
USING #Table2 T2
ON T1.gender = T2.gender
   AND T1.birthdate = T2.birthdate
WHEN MATCHED THEN
  UPDATE SET address = T2.address,
             phone2 = T2.phone; 

或者，如果有多个匹配项，则可以选择一个特定的记录

用 APPLY

UPDATE T1
SET    address = T2.address,
       phone2 = T2.phone
FROM   #Table1 T1
       CROSS APPLY (SELECT TOP 1 *
                    FROM   #Table2 T2
                    WHERE  T1.gender = T2.gender
                           AND T1.birthdate = T2.birthdate
                    ORDER  BY T2.PrimaryKey) T2 

..或CTE

WITH T2
     AS (SELECT *,
                ROW_NUMBER() OVER (PARTITION BY gender, birthdate ORDER BY primarykey) AS RN
         FROM   #Table2)
UPDATE T1
SET    address = T2.address,
       phone2 = T2.phone
FROM   #Table1 T1
       JOIN T2
         ON T1.gender = T2.gender
            AND T1.birthdate = T2.birthdate
            AND T2.RN = 1;

UPDATE TS
SET TS.TaskFullAddress = L.FullAddress
FROM [dbo].[TaskOrders]   TS
INNER JOIN Locations L
ON  TS.ClientId  = L.ClientId;