如何在 sql server 2008 中交换列值？

Question

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

我想将数据从AttributeValue列交换为AttributeName并将AttributeName交换为AttributeValue

例如：

Eno     ename     AttributeValue   AttributeName  
1       aa        abc              a123
2       bbb       dcf              b123
3       cc        wew3             c7sd

Answer 1

UPDATE employee
SET AttributeValue = AttributeName, 
    AttributeName = AttributeValue

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。

Answer 2

Update employee
Set attributeValue = attributeName,
    attributeName = attributeValue

Answer 3

update Employee set AttributeValue = AttributeName, AttributeName = AttributeValue

Answer 4

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";

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。

Answer 5

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;

Answer 6

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

输出

Answer 7

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)

Answer 8

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

更新员工 SET AttributeValuee= temp