SQL練習題_用戶購買收藏記錄合併（拼多多）

目錄

• 拼多多筆試題0805_統計用戶數據
  • 筆試題描述
  • 表格構建
    • 數據觀察
  • 題目分析
    • 1、合併表格
    • 2、CASE表示（0,1）
    • 3、同理複製FORK表
  • 題目解答

拼多多筆試題0805_統計用戶數據

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

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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');

表格結果以下：

TABLE buy

TABLE fork

數據觀察

• 表格中能夠發現以下問題

有些商品已購買，未收藏
有些商品未購買，已收藏

• 最後輸出中須要彙總全部用戶&商品

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題目分析

1、合併表格

• 當保證buy表全部數據時，應使用LEFT JOIN
• 如有數據有購買記錄，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;

表格結果以下：

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2、CASE表示（0,1）

• 根據是否爲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'

• 固然以buy爲主表，是不可能出現【未收藏已購買】【未收藏未購買】的狀況的。

最後結果及代碼

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

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3、同理複製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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題目解答

• 兩個結果合併後，便可獲得最終結果
• UNION - 去除重複行合併

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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