| * | * |
|---|---|
| 链接 | SQL Basics: Up and Down |
| 难度 | 7kyu |
| 状态 | √ |
| 日期 | 2019-4-17 |
题意
题解1
select
case when (number1 + number2) mod 2 == 0
then max(number1 , number2)
when (number1 + number2) mod 2 <> 0
then min(number1 , number2)
from numbers
题解2
select
case
when sum(number1) % 2 = 1 then min(number1)
else max(number1)
end as number1
,
case
when sum(number2) % 2 = 1 then min(number2)
else max(number2)
end as number2
from numbers
