SQL练习题_用户购买收藏记录合并（拼多多）
拼多多笔试题0805_统计用户数据
笔试题描述
本题来自2018年8月5日拼多多笔试题
- 给定两张表buy和fork分别记录用户的购买记录、收藏记录
- 返回状态“已收藏已购买”“已收藏未购买”“未收藏已购买”，以（0,1）表示
表格构建
create table buy(user_id int,item_id int,buy_time DATE);
create table fork(user_id int,item_id int ,fork_time DATE);
insert into buy values(0001,201,'2008-09-04');
insert into buy values(0001,206,'2008-09-04');
insert into buy values(0002,203,'2008-09-04');
insert into buy values(0003,204,'2008-09-04');
insert into fork values(0001,203,'2008-09-04');
insert into fork values(0001,201,'2008-09-04');
insert into fork values(0001,205,'2008-09-04');
insert into fork values(0004,203,'2008-09-04');
insert into fork values(0003,204,'2008-09-04');
insert into fork values(0002,201,'2008-09-04');
表格结果如下：
TABLE buy
TABLE fork
数据观察
- 表格中可以发现如下问题
有些商品已购买，未收藏
有些商品未购买，已收藏
- 最后输出中需要汇总所有用户&商品
题目分析
一、合并表格
- 当保证buy表所有数据时，应使用LEFT JOIN
- 若有数据有购买记录，无收藏记录，表格中则会显示NULL
SELECT *
FROM buy LEFT JOIN fork
ON buy.user_id=fork.user_id AND buy.item_id=fork.item_id;
表格结果如下：
二、CASE表示（0,1）
- 根据是否为NULL值，进行逻辑判断
CASE WHEN fork.fork_time is not null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'fork&buy',
CASE WHEN fork.fork_time is not null and buy.buy_time is null THEN 1 ELSE 0 END AS 'fork&NOT buy',
CASE WHEN fork.fork_time is null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'NOT fork&buy',
CASE WHEN fork.fork_time is null and buy.buy_time is null THEN 1 ELSE 0 END AS 'NOT fork& NOT buy'
- 当然以buy为主表，是不可能出现【未收藏已购买】【未收藏未购买】的情况的。
最后结果及代码
SELECT buy.user_id,buy.item_id,
CASE WHEN fork.fork_time is not null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'fork&buy',
CASE WHEN fork.fork_time is not null and buy.buy_time is null THEN 1 ELSE 0 END AS 'fork&NOT buy',
CASE WHEN fork.fork_time is null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'NOT fork&buy',
CASE WHEN fork.fork_time is null and buy.buy_time is null THEN 1 ELSE 0 END AS 'NOT fork& NOT buy'
FROM buy LEFT JOIN fork
ON buy.user_id=fork.user_id and buy.item_id=fork.item_id
三、同理复制FORK表
SELECT fork.user_id,fork.item_id,
CASE WHEN fork.fork_time is not null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'fork&buy',
CASE WHEN fork.fork_time is not null and buy.buy_time is null THEN 1 ELSE 0 END AS 'fork&NOT buy',
CASE WHEN fork.fork_time is null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'NOT fork&buy',
CASE WHEN fork.fork_time is null and buy.buy_time is null THEN 1 ELSE 0 END AS 'NOT fork& NOT buy'
FROM fork LEFT JOIN buy
ON buy.user_id=fork.user_id and buy.item_id=fork.item_id
题目解答
- 两个结果合并后，即可得到最终结果
- UNION - 去除重复行合并
SELECT buy.user_id,buy.item_id,
CASE WHEN fork.fork_time is not null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'fork&buy',
CASE WHEN fork.fork_time is not null and buy.buy_time is null THEN 1 ELSE 0 END AS 'fork&NOT buy',
CASE WHEN fork.fork_time is null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'NOT fork&buy',
CASE WHEN fork.fork_time is null and buy.buy_time is null THEN 1 ELSE 0 END AS 'NOT fork& NOT buy'
FROM buy LEFT JOIN fork
ON buy.user_id=fork.user_id and buy.item_id=fork.item_id
UNION
SELECT fork.user_id,fork.item_id,
CASE WHEN fork.fork_time is not null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'fork&buy',
CASE WHEN fork.fork_time is not null and buy.buy_time is null THEN 1 ELSE 0 END AS 'fork&NOT buy',
CASE WHEN fork.fork_time is null and buy.buy_time is not null THEN 1 ELSE 0 END AS 'NOT fork&buy',
CASE WHEN fork.fork_time is null and buy.buy_time is null THEN 1 ELSE 0 END AS 'NOT fork& NOT buy'
FROM fork LEFT JOIN buy
ON buy.user_id=fork.user_id and buy.item_id=fork.item_id
ORDER BY user_id,item_id;
