-- ① 運行報錯
SELECT a. Employee_id, DISTINCT b. Employee_id
FROM Employee a
LEFT JOIN Incentive b ON a. Employee_id = b. Employee_id
-- ② 運行正常
SELECT DISTINCT a. Employee_id, b. Employee_id
FROM Employee a
LEFT JOIN Incentive b ON a. Employee_id = b. Employee_id
DISTINCT 的位置不同導致的差異! ?②爲啥會報錯!暫且記住第二種方法