#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)
【SQL】sql update 多表关联更新方法总结
原文地址:https://www.cnblogs.com/ReAiQingYi/p/14865755.html