Question

考虑到一个带有id列和int列的简单MySQL表，我需要计算 gap 等于或大于某个值的次数。

让我们说价值将是10。

给出以下样本记录：

{1, 2, 3} = 1 time
{1, 2, 3, 4, 5, 6, 7, 8, 9} = 1 time;
{1, 2, 3, 14, 17} = 2 times (1, 2, 3 and 14, 17);
{1, 2, 3, 14, 20, 40, 42} = 3 times (1, 2, 3 and 14, 20 and 40, 42);

是否可以使用mysql解决问题？

Answer 1

是。对于列t和id的表num，这将是这样的：

SET @n = 10;

SELECT 1 + SUM(COALESCE(t3.f, 0))
FROM (
    SELECT DISTINCT t1.num, (
        SELECT CASE WHEN t2.num - t1.num > @n THEN 1 ELSE 0 END
        FROM t t2 
        WHERE t2.num > t1.num 
        ORDER BY num LIMIT 1
    ) AS f
    FROM t t1
) t3

当值之间的差距大于x

1 个答案: