-- ① 运行报错
SELECT a. Employee_id, DISTINCT b. Employee_id
FROM Employee a
LEFT JOIN Incentive b ON a. Employee_id = b. Employee_id
-- ② 运行正常
SELECT DISTINCT a. Employee_id, b. Employee_id
FROM Employee a
LEFT JOIN Incentive b ON a. Employee_id = b. Employee_id
DISTINCT 的位置不同导致的差异! ?②为啥会报错!暂且记住第二种方法