[LeetCode/MySQL50] 577. Employee Bonus
2025. 3. 10. 03:03ㆍ개인활동/코테
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Problem
Solution
select E.name, B.bonus
from Employee as E
left join Bonus as B
on E.empID = B.empID
where B.bonus < 1000 or B.bonus is null;
Null값도 포함 시켜야 하기 때문에 where 절에서 is null을 활용해주기
맨날 이 부분을 까먹는 것 같다
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