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数据说明：表名：order字段名：userid、paytime查询要求：去重查询每日下单人数，即同一个人同一天下N单，只计算为1个下单人数。 SELECT "count"(1),date1 FROM (SELECT user_id,to_char(pay_time, 'yyyy-MM-dd') date1 FROM obs_order WHERE pay_time BETWEEN '2021-07-31 23:59:59' AND '2021-08-04 23:59:59' AND statu.

数据说明：表名：user字段名：opendate，类型：varcharSELECT "count"(1) FROM "user" WHERE opendate BETWEEN to_char(CURRENT_DATE - 2, 'yyyy-mm-dd') AND to_char(CURRENT_DATE, 'yyyy-mm-dd');详细说明：1.count(1)：查询记录总数量；2.to_char(CURRENT_DATE - 2, ‘yyyy-mm-dd’)：将前天日期转换成varcha

表Weather:idrecordDateTemperature12015-01-011012015-01-022512015-01-032012015-01-0430思路分析：使用MySQL DATEDIFF() 函数MySQL语句：select w1.id as ‘Id’ from Weather w1 join Weather w2 on DATEDIFF(w1.recordDate,w2.recordDate) = 1 and w

Person表：IdEmail1a@163.com2b@163.com3a@163.com期望结果：IdEmail1a@163.com2b@163.com思路分析：①从期望结果来看，直达的效果就是将重复的记录删除；②首先，需要找到那些重复的记录。我们可以将Person表作为p1表和p2表进行连接，先找到重复的记录。select p1.* from Person p1,Person p2 where p1.Emali =

查询第二高薪水表名EmployeeIdSaraly110022003300预期结果：SecondHighestSalary200MySQL：select max(Salary) as SecondHighestSalary from (select Salary from Employee where Salary < (select max(Salary) from Employee)) a;思路分析：第二高薪水，即去