[英]How do I swap column values in sql server 2008?
I have a table called Employee我有一张名为 Employee 的表
Eno ename AttributeValue AttributeName
1 aa a123 abc
2 bbb b123 dcf
3 cc c7sd wew3
I want to swap the data from column AttributeValue
to AttributeName
and AttributeName
to AttributeValue
我想将数据从
AttributeValue
列交换为AttributeName
并将AttributeName
交换为AttributeValue
For Example:例如:
Eno ename AttributeValue AttributeName
1 aa abc a123
2 bbb dcf b123
3 cc wew3 c7sd
UPDATE employee
SET AttributeValue = AttributeName,
AttributeName = AttributeValue
However, unless both columns have the exact same definition, you risk losing information.但是,除非两列具有完全相同的定义,否则您可能会丢失信息。
Update employee
Set attributeValue = attributeName,
attributeName = attributeValue
update Employee set AttributeValue = AttributeName, AttributeName = AttributeValue
All the previous techniques are slow for big tables they move data instead of renaming columns, this is a simple solution:所有以前的技术对于大表来说都很慢,它们移动数据而不是重命名列,这是一个简单的解决方案:
ALTER TABLE "amplitude"
RENAME COLUMN "start_hour_displayed" TO "temp";
ALTER TABLE "amplitude"
RENAME COLUMN "start_hour" TO "start_hour_displayed";
ALTER TABLE "amplitude"
RENAME COLUMN "temp" TO "start_hour";
If you have views of functions linked you have to backup them before and restore them after.如果您有链接的功能视图,您必须先备份它们,然后再恢复它们。
This is really good example这真是个好例子
SELECT * from employees;
Go
DECLARE @temp as varchar(20)
update employees
set @temp = fname,
fname = lname,
lname = @temp
WHERE deptno = 10;
GO
SELECT * from employees;
Declare @myTable Table (id int, first_name varchar(50), last_name varchar(50));
Select * from Student
Insert Into @myTable (id, first_name, last_name) Select id, last_name, first_name from Student
MERGE
INTO Student std
USING @myTable tmp
ON std.id = tmp.id
WHEN MATCHED THEN
UPDATE
SET std.first_name = tmp.first_name,
std.last_name = tmp.last_name;
Select * from Student
Output输出
Just swap both columns in a single update:只需在一次更新中交换两列:
Update registration
Set AttributeName = AttributeValue ,
AttributeValue = AttributeName where id in (1,2,3)
UPDATE employee SET AttributeValue = AttributeName, AttributeName = AttributeValue its not correct better to create one temp column and try the code UPDATE employee SET temp= AttributeName then again UPDATE employee SET AttributeName = AttributeValuee again更新员工 SET AttributeValue = AttributeName, AttributeName = AttributeValue 创建一个临时列并尝试代码更新员工 SET temp= AttributeName 然后再次更新员工 SET AttributeName = AttributeValuee 是不正确的
UPDATE employee SET AttributeValuee= temp更新员工 SET AttributeValuee= temp
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