X 市建了一个新的体育馆，每日人流量信息被记录在这三列信息中：序号 (id)、日期 (visit_date)、 人流量 (people)。
请编写一个查询语句，找出人流量的高峰期。高峰期时，至少连续三行记录中的人流量不少于100。
例如，表 stadium：
+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 1 | 2017-01-01 | 10 |
| 2 | 2017-01-02 | 109 |
| 3 | 2017-01-03 | 150 |
| 4 | 2017-01-04 | 99 |
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
对于上面的示例数据，输出为：
+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
提示：
每天只有一行记录，日期随着 id 的增加而增加。
题目来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/human-traffic-of-stadium
审题：查询people连续超过100三天以上的所有。
思考：
解题：
解法一
先将哪些连续三天客流量都>=100的行找出来。
表三次连接。结果命名为表A。
(
SELECT S1.id AS `id1`,S2.id AS `id2`,S3.id AS `id3`
FROM stadium AS S1
-- 两天大于100
JOIN stadium AS S2 ON(S1.id +1 = S2.id AND S1.people >=100 AND S2.people >=100)
-- 三天大于
JOIN stadium AS S3 ON(S2.id +1 = S3.id AND S2.people >=100 AND S3.people >=100)
) AS A那么，id1，id2，id3都是满足条件的行号。
与表Stadium连接。取出id1或id2或id3对应的行，但可能有些行重复取，要去重。
SELECT DISTINCT S.*
FROM
(
SELECT S1.id AS `id1`,S2.id AS `id2`,S3.id AS `id3`
FROM stadium AS S1
JOIN stadium AS S2 ON(S1.id +1 = S2.id AND S1.people >=100 AND S2.people >=100)
JOIN stadium AS S3 ON(S2.id +1 = S3.id AND S2.people >=100 AND S3.people >=100)
) AS A
JOIN stadium AS S ON(A.id1 = S.id OR A.id2=S.id OR A.id3=S.id)解法二
对每一行A，如果A的前两行的客流量都>=100，或者A的前一行或后一行的客流量都>=100，或者 A的后两行的客流量都>=100 。那么，行A就是结果中的一条。
-- 每次查找一条数据
SELECT *
FROM stadium AS S1
WHERE S1.people >= 100 AND
(
2=(
SELECT COUNT(*)
FROM stadium AS S
WHERE S.people >= 100 AND (S.id = S1.id-1 OR S.id = S1.id-2)
) OR
2=(
SELECT COUNT(*)
FROM stadium AS S
WHERE S.people >= 100 AND (S.id = S1.id-1 OR S.id = S1.id+1)
) OR
2=(
SELECT COUNT(*)
FROM stadium AS S
WHERE S.people >= 100 AND (S.id = S1.id+1 OR S.id = S1.id+2)
)
)A的前两行的客流量都>=100，逻辑如下：
2=(
SELECT COUNT(*)
FROM stadium AS S
WHERE S.people >= 100 AND (S.id = S1.id-1 OR S.id = S1.id-2)
)A的前一行或后一行的客流量都>=100，逻辑如下：
2=(
SELECT COUNT(*)
FROM stadium AS S
WHERE S.people >= 100 AND (S.id = S1.id-1 OR S.id = S1.id+1)
)A的后两行的客流量都>=100，逻辑如下：
2=(
SELECT COUNT(*)
FROM stadium AS S
WHERE S.people >= 100 AND (S.id = S1.id+1 OR S.id = S1.id+2)
)解法三
与解法一思路相近。
表自叉积三次。
筛选出客流量>=100，并且三个id依次加1的行。
SELECT DISTINCT S1.*
FROM stadium AS S1,stadium AS S2,stadium AS S3
WHERE S1.people>=100 AND S2.people>=100 AND S3.people>=100 AND (
-- 后两天
S1.id +1 = S2.id AND S1.id+2=S3.id OR
-- 一前一后
S1.id +1 = S2.id AND S1.id-1=S3.id OR
-- 前两天
S1.id -1 = S2.id AND S1.id-2=S3.id
)
-- 排序
ORDER BY S1.id知识点：
