https://www.heckjj.com/index.php zh-cn https://www.heckjj.com/post// Heck <@hecks.tk> Fri, 02 Jul 2021 13:11:37 +0000 https://www.heckjj.com/post// 1、表一：Test1
Id  name  age
1      
2      

2、表二：Test2
Id  name  age
1  小明  10
2  小红  8


#实现将表Test2的name和age字段数据更新到表Test1中，按照id相等的条件

1、SQLServer多表更新方法：
语法：

UPDATE { table_name WITH ( < table_hint_limited > [ ...n ] ) &#124; view_name &#124; rowset_function_limited }
SET { column_name = { expression &#124; DEFAULT &#124; NULL } &#124; @variable = expression &#124; @variable = column = expression } [ ,...n ]
{ { [ FROM { < table_source > } [ ,...n ] ] [ WHERE < search_condition > ] } &#124; [ WHERE CURRENT OF { { [ GLOBAL ] cursor_name } &#124; cursor_variable_name } ] } [ OPTION ( < query_hint > [ ,...n ] ) ]
例子：

update test1
set test1.name=test2.name,test1.age=test2.age
from test1
inner join test2
on test1.id=test2.id

2、Oracle 多表更新方法：
语法：

UPDATE updatedtable
SET (col_name1[,col_name2...])= (SELECT col_name1,[,col_name2...]
FROM srctable [WHERE where_definition])
例子：

update test1
set (test1.name,test1.age)=
(select test2.name,test2.age from test2 where test2.id=test1.id)

3、MySql多表更新方法：
语法：

UPDATE table_references
SET col_name1=expr1 [, col_name2=expr2 ...] [WHERE where_definition]
例子：

update test1,test2
set test1.name=test2.name,test1.age=test2.age
where test1.id=test2.id

4、通用方法：(*^__^*)
update test1
set name=(select name from test2 where test2.id=test1.id),
age=(select age from test2 where test2.id=test1.id) ]]> https://www.heckjj.com/post//#blogcomment <user@domain.com> Thu, 01 Jan 1970 00:00:00 +0000 https://www.heckjj.com/post//#blogcomment