SQL练习题_用户购买收藏记录合并（拼多多）

目录

• 拼多多笔试题0805_统计用户数据
  • 笔试题描述
  • 表格构建
    • 数据观察
  • 题目分析
    • 1、合并表格
    • 2、CASE表示（0,1）
    • 3、同理复制FORK表
  • 题目解答

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

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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');

表格结果以下：

TABLE buy

TABLE fork

数据观察

• 表格中能够发现以下问题

有些商品已购买，未收藏
有些商品未购买，已收藏

• 最后输出中须要汇总全部用户&商品

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题目分析

1、合并表格

• 当保证buy表全部数据时，应使用LEFT JOIN
• 如有数据有购买记录，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;

表格结果以下：

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2、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'

• 固然以buy为主表，是不可能出现【未收藏已购买】【未收藏未购买】的状况的。

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

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3、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

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题目解答

• 两个结果合并后，便可获得最终结果
• 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;

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