mysql-关于最大值的内部联接表
作者:互联网
我试图编写一个MySQL查询,在其中提取卖方的信息和她最受欢迎的产品.这取决于浏览量最多的产品,即MAX(page_views).
不过,以下查询只是拉出一种随机产品,而不是浏览量最多的产品.
"SELECT
seller.id, seller.language, seller.shop_name,seller.story,
seller.eng_story, product.id, product.image_thumb, product.title,
product.eng_title, product.price, MAX(product.page_views)
FROM seller
INNER JOIN product ON seller.id=product.seller_id
WHERE seller.handpicked='y' AND seller.shop_active='y'
GROUP BY seller.id
ORDER BY product.page_views
LIMIT 0,5"
或更准确地说,page_views实际上是正确的数字,但是如何获得其他产品字段(id,图片,标题等),以使其具有最多的页面浏览量.
表数据:
卖家:
id | language | shop_Name | story | eng_story | handpicked | active
1 | 1 | mitienda | hola mundo| Hello world | Y | Y
2 | 1 | sisenor | bonita | beautiful | N | Y
3 | 2 | new_world | mi vida | my life | Y | Y
产品:
id | seller_id | image_thumb | title | eng_title | price | page Views
1 | 1 | /images/.. | sombrero | hat | $5 | 10
2 | 1 | /images/.. | bufanda | scarf | $25 | 30
3 | 2 | /images/.. | arte | art | $15 | 15
4 | 3 | /images/.. | joyeria | jewlery | $10 | 1
5 | 2 | /images/.. | canasta | basket | $21 | 13
6 | 3 | /images/.. | ropa | clothes | $13 | 6
预期结果(精简):
seller.id | shop_name | product.id | pageviews | title | price
1 | miteinda | 2 | 30 | bufanda | $25
3 | newworld | 6 | 6 | ropa | $13
结果应按浏览量列出经过手工挑选的卖家信息及其最受欢迎的产品.按浏览量对卖家进行排序,总共限制为5个卖家.
解决方法:
SELECT a.ID SellerID,
a.Shop_Name,
b.ID ProductID,
b.pageViews,
b.title,
b.Price
FROM seller a
INNER JOIN Products b
ON a.id = b.seller_ID
INNER JOIN
(
SELECT seller_ID, MAX(pageViews) max_view
FROM products
GROUP BY seller_ID
) c ON b.seller_ID = c.seller_ID AND
b.pageViews = c.max_View
WHERE a.handpicked = 'Y' AND a.active = 'Y'
OUTPUT
╔══════════╦═══════════╦═══════════╦═══════════╦═════════╦═══════╗
║ SELLERID ║ SHOP_NAME ║ PRODUCTID ║ PAGEVIEWS ║ TITLE ║ PRICE ║
╠══════════╬═══════════╬═══════════╬═══════════╬═════════╬═══════╣
║ 1 ║ mitienda ║ 2 ║ 30 ║ bufanda ║ $25 ║
║ 3 ║ new_world ║ 6 ║ 6 ║ ropa ║ $13 ║
╚══════════╩═══════════╩═══════════╩═══════════╩═════════╩═══════╝
标签:sql,mysql,select,greatest-n-per-group 来源: https://codeday.me/bug/20191010/1888251.html