Leetcode1083. 銷售分析II（簡單）

Table: Product

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| product_id   | int     |
| product_name | varchar |
| unit_price   | int     |
+--------------+---------+

product_id 是這張表的主鍵

Table: Sales

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| seller_id   | int     |
| product_id  | int     |
| buyer_id    | int     |
| sale_date   | date    |
| quantity    | int     |
| price       | int     |
+------ ------+---------+

這個表沒有主鍵衣撬，它可以有重復(fù)的行.
product_id 是 Product 表的外鍵.
編寫一個 SQL 查詢摩疑，查詢購買了 S8 手機(jī)卻沒有購買 iPhone 的買家。注意這里 S8 和 iPhone 是 Product 表中的產(chǎn)品嵌器。

查詢結(jié)果格式如下圖表示：
Product table:

+------------+--------------+------------+
| product_id | product_name | unit_price |
+------------+--------------+------------+
| 1          | S8           | 1000       |
| 2          | G4           | 800        |
| 3          | iPhone       | 1400       |
+------------+--------------+------------+

Sales table:

+-----------+------------+----------+------------+----------+-------+
| seller_id | product_id | buyer_id | sale_date  | quantity | price |
+-----------+------------+----------+------------+----------+-------+
| 1         | 1          | 1        | 2019-01-21 | 2        | 2000  |
| 1         | 2          | 2        | 2019-02-17 | 1        | 800   |
| 2         | 1          | 3        | 2019-06-02 | 1        | 800   |
| 3         | 3          | 3        | 2019-05-13 | 2        | 2800  |
+-----------+------------+----------+------------+----------+-------+

Result table:

+-------------+
| buyer_id    |
+-------------+
| 1           |
+-------------+

id 為 1 的買家購買了一部 S8粗梭，但是卻沒有購買 iPhone并鸵，而 id 為 3 的買家卻同時購買了這 2 部手機(jī)。

解答
想法是創(chuàng)建購買了S8的臨時表再創(chuàng)建沒有購買iphone的臨時表兩表連接即可

先選出 S8 iphone的id

select product_id
from Product
where product_name = 'S8';

select product_id
from Product
where product_name = 'iPhone';

選出購買了S8的買家

select buyer_id
from Sales 
where product_id = (select product_id
from Product
where product_name = 'S8')

選出沒有購買iphone的id

select buyer_id
from Sales 
where product_id <> (select product_id
from Product
where product_name = 'iPhone')

兩表連接

SELECT tmp.buyer_id AS buyer_id
FROM (SELECT buyer_id
FROM Sales 
WHERE product_id = (SELECT product_id
FROM Product
WHERE product_name = 'S8')) tmp
JOIN (SELECT buyer_id
FROM Sales 
WHERE product_id <> (SELECT product_id
FROM Product
WHERE product_name = 'iPhone')) tmp2
ON tmp.buyer_id = tmp2.buyer_id

也可以轉(zhuǎn)為集合差的問題從買了S8的集合中刪去買了iphone的集合

select A.buyer_id
from
(
    select distinct buyer_id
    from Product as P join Sales as S
        on(P.product_id = S.product_id and P.product_name ='S8')
) as A
left join
(
    select distinct buyer_id
    from Product as P join Sales as S
        on(P.product_id = S.product_id and P.product_name ='iPhone')
) as B
    on(A.buyer_id = B.buyer_id)
where B.buyer_id is NULL

別的解答
先做兩表連接

SELECT *
FROM Sales AS S LEFT JOIN Product AS P 
    ON(P.product_id = S.product_id)

對每個買家統(tǒng)計購買S8和Iphone的數(shù)量

SELECT SUM(IF(P.`product_name` = 'S8', 1, 0)), SUM(IF(P.`product_name` = 'iPhone', 1, 0)) 
FROM Sales AS S 
LEFT JOIN Product AS P 
ON(P.product_id = S.product_id)
GROUP BY S.`seller_id`

再選出購買S8的數(shù)量>0 而購買iPhone數(shù)量=0的買家id即可

SELECT S.`buyer_id`
FROM Sales AS S 
LEFT JOIN Product AS P 
ON(P.product_id = S.product_id)
GROUP BY S.`seller_id`
HAVING SUM(IF(P.`product_name` = 'S8', 1, 0)) > 0 AND  SUM(IF(P.`product_name` = 'iPhone', 1, 0)) = 0

最后編輯于：2020.07.26 16:59:28

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