Question

SELECT DISTINCT(id_no), lastname,
    (SELECT COUNT(purchasedate)  num_of_purch
    FROM sales JOIN Artist ON
        sales.id = Artist.id_no
        WHERE  DATE_SUB(CURDATE(),INTERVAL 1
                          YEAR) <= purchasedate 
              ) AS num_of_purch

FROM Artist

此查询返回所有艺术家的ID_no，他们的姓氏和购买总数，我想指定哪些购买是哪个艺术家。有助于解决这个问题。

编辑 - DISTINCT（id_no）是多余的，因为它是主键。

Answer 1

显示每个artist_id的销售数量：

SELECT artist.id_no, count(sales.id) as num_of_purch
FROM artist left join sales on sales.id = artist.id_no
WHERE DATE_SUB(CURDATE(), INTERVAL 1 YEAR) <= purchasedate
GROUP BY artist.id

还要返回姓氏和所有细节：

SELECT art_tot.id_no, art_tot.lastname, art_tot.num_of_purch, sales.*
FROM (SELECT artist.id_no, artist.lastname, count(sales.id) as num_of_purch
      FROM artist left join sales on sales.id = artist.id_no
      WHERE DATE_SUB(CURDATE(), INTERVAL 1 YEAR) <= purchasedate
      GOUP BY artist.id, artist.lastname) art_tot
      left join sales on art_tot.id_no = sales.id

Answer 2

这应该为您提供艺术家和每位艺术家的购买数量

select a.id_no, a.lastname, count(s.purchasedate) num_of_purch
from artists a
join sales s on a.id_no = s.id
where date_sub(curdate(), interval 1 year) <= s.purchasedate
group by a.id_no, a.lastname

Answer 3

您应该使用GROUP BY来获取每个艺术家的计数。

你应该使用外部联接来包括去年没有销售的艺术家。

SELECT a.id_no, a.lastname, COUNT(s.purchasedate) AS num_of_purch
FROM Artist a
LEFT OUTER JOIN sales s ON s.id = a.id_no
  AND s.purchasedate => CURDATE() - INTERVAL 1 YEAR
GROUP BY a.id_no;

PS：使用DISTINCT(id_no)没有意义，不仅因为id_no已经是唯一键，而且因为DISTINCT始终适用于选择列表中的所有列，即使您添加了括号来制作它看起来像一个仅适用于一列的函数。

MYSQL - 子查询问题

3 个答案: