Mysql 使用未来年份创建@X行

我正在尝试创建一个SQL，它将显示未来年份的@X行：

year(now())+1
year(now())+2
year(now())+3
...

@X是用户定义的变量

我只想使用union select，但它看起来很糟糕，适用于相当低的@x:

set @x:=3;
select year(now())+1 as year
union select if(@x>1, year(now())+2,null) as year
union select if(@x>2, year(now())+3,null) as year
union select if(@x>3, year(now())+4,null) as year;

---

这将是自MySQL 8以来可用的递归CTE的领域

SET @x = 5;

WITH RECURSIVE
years
AS
(
SELECT year(now()) year
UNION ALL
SELECT year + 1
       FROM years
       WHERE year + 1 - year(now()) <= @x
)
SELECT *
       FROM years;