SQL每日一题(20200513)

引自:https://mp.weixin.qq.com/s?__biz=MzA3MTg4NjY4Mw==&mid=2457305432&idx=4&sn=f945fef1267983d9aad405d2b1397c22&chksm=88a5936cbfd21a7a901ab9a764285477a36987a1a17193dd194025d4e11f962dcf0e5580377a&mpshare=1&scene=1&srcid=&sharer_sharetime=1589379613101&sharer_shareid=18c156b37f741bf9989098e28bf33e09&exportkey=Abczad4uJ2LGc0e82s%2FlStE%3D&pass_ticket=W3S3KdK%2BnUpk0DsaHX5cmRIsPCVEMCdRjXKBld%2Bn9Ffjas8N%2BjYwZWCDsVPti4EQ#rd

感谢出题者,如有冒犯,请与我联系,希望和大家一起学习交流。(目前使用oracle数据库环境编写sql)

 

 

题目

有一个商场,每日人流量信息被记录在这三列信息中:序号 (id)、日期 (date)、 人流量 (people)。请编写一个查询语句,找出高峰期时段,要求连续三天及以上,并且每天人流量均不少于100。例如,表 stadium:
 

 

 

    对于上面的示例数据,输出为:

 

 

该如何写这个SQL?

 

读者可以试着自己思考写下,再往下翻... 如有不同解题方式,大家一起交流。

 

 

 

 

 

ID这边当作是连续有序值,如果 实际环境中不是的,以下sql需要再嵌一层,列出有序列代替ID:

sys@WIL> WITH T AS 2 (SELECT 1 ID, DATE 2019-01-01 DT, 10 PEOPLE 3 FROM DUAL 4 UNION ALL 5 SELECT 2 ID, DATE 2019-01-02 DT, 109 PEOPLE 6 FROM DUAL 7 UNION ALL 8 SELECT 3 ID, DATE 2019-01-03 DT, 150 PEOPLE 9 FROM DUAL 10 UNION ALL 11 SELECT 4 ID, DATE 2019-01-04 DT, 99 PEOPLE 12 FROM DUAL 13 UNION ALL 14 SELECT 5 ID, DATE 2019-01-05 DT, 145 PEOPLE 15 FROM DUAL 16 UNION ALL 17 SELECT 6 ID, DATE 2019-01-06 DT, 1455 PEOPLE 18 FROM DUAL 19 UNION ALL 20 SELECT 7 ID, DATE 2019-01-07 DT, 199 PEOPLE 21 FROM DUAL 22 UNION ALL 23 SELECT 8 ID, DATE 2019-01-08 DT, 188 PEOPLE 24 FROM DUAL) 25 SELECT ID, DT, PEOPLE 26 FROM (SELECT TT.*, COUNT(*) OVER(PARTITION BY GID) CN 27 FROM (SELECT T.*, ID - ROW_NUMBER() OVER(ORDER BY DT) GID 28 FROM T 29 WHERE PEOPLE > 100) TT) 30 WHERE CN >= 3 31 ORDER BY 2; ID DT PEOPLE---------- ---------- ---------- 5 2019-01-05 145 6 2019-01-06 1455 7 2019-01-07 199 8 2019-01-08 188

 

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