拼多多筆試題0805_統計用戶數據
目錄
筆試題描述
本題來自2018年8月5日拼多多筆試題
- 給定兩張表buy和fork分別記錄用戶的購買記錄、收藏記錄
- 返回狀態“已收藏已購買”“已收藏未購買”“未收藏已購買”，以（0,1）表示
表格構建
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');
表格結果如下：
TABLE buy
TABLE fork
數據觀察
- 表格中可以發現如下問題
有些商品已購買，未收藏
有些商品未購買，已收藏
- 最后輸出中需要匯總所有用戶&商品
題目分析
一、合並表格
- 當保證buy表所有數據時，應使用LEFT JOIN
- 若有數據有購買記錄，無收藏記錄，表格中則會顯示NULL
SELECT *
FROM buy LEFT JOIN fork
ON buy.user_id=fork.user_id AND buy.item_id=fork.item_id;
表格結果如下：
二、CASE表示（0,1）
- 根據是否為NULL值，進行邏輯判斷
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'
- 當然以buy為主表，是不可能出現【未收藏已購買】【未收藏未購買】的情況的。
最后結果及代碼
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、同理復制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
題目解答
- 兩個結果合並后，即可得到最終結果
- UNION - 去除重復行合並
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;
