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【SQL】sql update 多表关联更新方法总结

程序员文章站 2022-06-01 16:46:04
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#表结构:

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 ] ) | view_name | rowset_function_limited } 
SET { column_name = { expression | DEFAULT | NULL } | @variable = expression | @variable = column = expression } [ ,...n ] 
{ { [ FROM { < table_source > } [ ,...n ] ] [ WHERE < search_condition > ] } | [ WHERE CURRENT OF { { [ GLOBAL ] cursor_name } | 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

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@友情链接:https://www.cnblogs.com/willingtolove/p/10792713.html