如何在 sql server 2008 中交換列值？

Question

我有一張名為 Employee 的表

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

我想將數據從AttributeValue列交換為AttributeName並將AttributeName交換為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，除非兩列具有完全相同的定義，否則您可能會丟失信息。

Answer 2

Update employee
Set attributeValue = attributeName,
    attributeName = attributeValue

Answer 3

update Employee set AttributeValue = AttributeName, AttributeName = AttributeValue

Answer 4

所有以前的技術對於大表來說都很慢，它們移動數據而不是重命名列，這是一個簡單的解決方案：

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";

如果您有鏈接的功能視圖，您必須先備份它們，然后再恢復它們。

Answer 5

這真是個好例子

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

輸出

Answer 7

只需在一次更新中交換兩列：

Update registration
Set AttributeName = AttributeValue ,
    AttributeValue = AttributeName where id in (1,2,3)

Answer 8

更新員工 SET AttributeValue = AttributeName, AttributeName = AttributeValue 創建一個臨時列並嘗試代碼更新員工 SET temp= AttributeName 然后再次更新員工 SET AttributeName = AttributeValuee 是不正確的

更新員工 SET AttributeValuee= temp