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牛客257734998号
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University of Chicago
2022
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IP属地:广东
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2022-04-11 21:41
University of Chicago 产品经理
题解 | #插入记录(三)#
replace into examination_info (exam_id,tag,difficulty,duration,release_time) values (9003,'SQL','hard',90,'2021-01-01 00:00:00 ') //replace into 和 insert into 的区别?
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2022-04-11 20:45
University of Chicago 产品经理
题解 | #插入记录(二)#
insert into exam_record_before_2021(uid,exam_id,start_time,submit_time,score) select uid,exam_id,start_time,submit_time,score from exam_record where submit_time < '2021-01-01 00:00:00'
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2022-04-11 20:13
University of Chicago 产品经理
题解 | #插入记录(一)#
insert into exam_record (uid,exam_id,start_time,submit_time,score ) values (1001,9001,'2021-09-01 22:11:12','2021-09-01 23:01:12','90'), (1002,9002,'2021-09-04 07:01:02',null,null)
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2022-04-11 17:29
University of Chicago 产品经理
题解 | #检索每个顾客的名称和所有的订单号(二)#
select Customers.cust_name,Orders.order_num from Orders right join Customers on Orders.cust_id = Customers.cust_id order by cust_name
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2022-04-11 17:13
University of Chicago 产品经理
题解 | #确定最佳顾客的另一种方式(二)#
select cust_name, sum(item_price * quantity) as total_price from OrderItems inner join Orders on OrderItems.order_num = Orders.order_num inner join Customers on Orders.cust_id = Customers.cust_id group by Customers.cust_name having total_price >= 1000 order by total_price
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2022-04-11 16:48
University of Chicago 产品经理
题解 | #返回购买 prod_id 为 BR01 的产品的所有顾客的电子邮件(二)#
select cust_email from Customers inner join Orders on Orders.cust_id = Customers.cust_id inner join OrderItems on OrderItems.order_num = Orders.order_num where prod_id = 'BR01'
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2022-04-08 20:20
University of Chicago 产品经理
题解 | #确定哪些订单购买了 prod_id 为 BR01 的产品(二)#
select a.cust_id,a.order_date from (select OrderItems.prod_id,Orders.cust_id,Orders.order_date from OrderItems inner join Orders on OrderItems.order_num = Orders.order_num where OrderItems.prod_id = 'BR01') as a order by a.order_date //子查询的用法
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2022-04-08 19:51
University of Chicago 产品经理
题解 | #返回顾客名称和相关订单号以及每个订单的总价#
select cust_name, Orders.order_num,sum(quantity * item_price) as OrderTotal from Customers inner join Orders on Customers.cust_id = Orders.cust_id inner join OrderItems on Orders.order_num = OrderItems.order_num group by Customers.cust_name,Orders.order_num \\为什呢这里一定要group by,题目也没说要啊?? order by Cus...
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2022-04-08 18:24
University of Chicago 产品经理
题解 | #从 Products 表中检索所有的产品名称以及对应的销售总数#
select prod_name,sum(quantity) as quant_sold from OrderItems inner join Products on Products.prod_id = OrderItems.prod_id group by prod_name
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2022-04-01 16:51
University of Chicago 产品经理
题解 | #找出每个学校GPA最低的同学#
select a.device_id,a.university,a.gpa from (select device_id,university,gpa, rank()over(partition by university order by gpa) as rank1 // 为什么不能将列名称命名为rank........ from user_profile) as a where a.rank1 = 1 //子查询里 按照学校分类,gpa排名由小到大排名;复查询里取各个分类中gpa排名最小的就是最低的gpa order by university
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2022-03-31 09:35
University of Chicago 产品经理
题解 | #21年8月份练题总数#
select count(distinct device_id) as did_cnt, count(question_id)as question_cnt from question_practice_detail where year(date)= 2021 and month(date) = 8 // where 的内容要用 and 连接
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2022-03-30 16:47
University of Chicago 产品经理
题解 | #提取博客URL中的用户名#
select device_id,substring_index(blog_url,'/',-1) as user_name from user_submit // select a,b 的顺序 就是新表的表头列的顺序 // 可以自己去看分隔符,用哪个分隔符取的数最方便
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2022-03-30 16:43
University of Chicago 产品经理
题解 | #截取出年龄#
select substring_index(substring_index(profile, ',',3), ',', -1) as age, //先截取出(a,b,c),再截取出 c count(device_id) as number from user_submit group by age //注意很多次写对了但不通过都是因为用了中文的括号
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2022-03-30 16:31
University of Chicago 产品经理
题解 | #统计每种性别的人数#
select substring_index(profile,',',-1) as gender, count(device_id)as number //substring_index(字段名称,'分隔符',第几位) from user_submit group by gender
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2022-03-30 15:43
University of Chicago 产品经理
题解 | #计算用户8月每天的练题数量#
select day(date) as day,count(question_id) as question_cnt // day (date)意思是取这个这天的日子作为新表中day的字段取值 from question_practice_detail where year(date)='2021' and month(date)='08' //意思是取其中年份是2021,月份是08的 group by day
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