php – 带有条件的Mysql案例

如何在 mysql查询中编写一个案例,该案例检查特定列的null或0

CREATE TABLE tblConfirmationStatus (Confirm_Status TINY INT)

INSERT INTO tblConfirmationStatus 
Confirm_Status
VALUES
(1),(0),(1),({null}),({null})

要求的输出

ConfirmStatus

   Confirmed
   Not Confirmed
   Confirmed
   Not Confirmed
   Not Confirmed
   Confirmed
   Not Confirmed

0或Null – 未确认,1 – 确认

SELECT CASE Confirm_Status 
            WHEN NULL OR 0 THEN 'Not Confirmed' 
             ELSE  'Confirmed' END AS ConfirmStatus
  FROM tblConfirmationStatus;

CASE语句有两种选择 – 您发布的语句或：

SELECT CASE 
        WHEN Confirm_Status IS NULL OR Confirm_Status = 0 THEN 'Not Confirmed' 
        ELSE  'Confirmed' 
       END AS ConfirmStatus

但你可以使用：

SELECT CASE 
        WHEN Confirm_Status > 0 THEN 'Confirmed' 
        ELSE  'Not Confirmed' 
       END AS ConfirmStatus

NULL是缺少值,因此检查大于零的值应该归入与零相同的类别.