2023-01-11：体育馆的人流量。编写一个 SQL 查询以找出每行的人数大于或等于 100 且 id 连续的三行或更多行记录。返回按 visit_date 升序排列的结果表。 DROP TAB

2023-01-11
北京
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2023-01-11：体育馆的人流量。编写一个 SQL 查询以找出每行的人数大于或等于 100 且 id 连续的三行或更多行记录。返回按 visit_date 升序排列的结果表。

DROP TABLE IF EXISTS `stadium`;CREATE TABLE `stadium` (  `id` int(11) NOT NULL,  `visit_date` date NOT NULL,  `people` int(11) NOT NULL,  PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=latin1;
INSERT INTO `stadium` VALUES ('1', '2017-01-01', '10');INSERT INTO `stadium` VALUES ('2', '2017-01-02', '109');INSERT INTO `stadium` VALUES ('3', '2017-01-03', '150');INSERT INTO `stadium` VALUES ('4', '2017-01-04', '99');INSERT INTO `stadium` VALUES ('5', '2017-01-05', '145');INSERT INTO `stadium` VALUES ('6', '2017-01-06', '1455');INSERT INTO `stadium` VALUES ('7', '2017-01-07', '199');INSERT INTO `stadium` VALUES ('8', '2017-01-09', '188');

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答案 2023-01-11：

sql 语句如下：

select distinct t1.*from stadium t1, stadium t2, stadium t3where t1.people >= 100 and t2.people >= 100 and t3.people >= 100and(    (t1.id - t2.id = 1 and t1.id - t3.id = 2 and t2.id - t3.id =1)  -- t1, t2, t3    or    (t2.id - t1.id = 1 and t2.id - t3.id = 2 and t1.id - t3.id =1) -- t2, t1, t3    or    (t3.id - t2.id = 1 and t2.id - t1.id =1 and t3.id - t1.id = 2) -- t3, t2, t1)order by t1.id