我正在尝试编写MySQL查询,其中我提取卖家的信息和她最受欢迎的产品。这取决于具有最多页面浏览量的产品,即MAX(page_views)
。
下面的查询只是拉随机产品而不是页面浏览量最多的产品。
"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"
或者更好地说,page_views
实际上是正确的数字,但我如何获得其他产品字段(ID,图片,标题等)页面浏览次数最多的产品。
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(简明):
seller.id | shop_name | product.id | pageviews | title | price
1 | miteinda | 2 | 30 | bufanda | $25
3 | newworld | 6 | 6 | ropa | $13
结果应列出已经过精选的卖家信息以及通过浏览量查看最受欢迎的产品。卖家按网页浏览排序,总共限制5个卖家。
答案 0 :(得分:13)
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'
输出
╔══════════╦═══════════╦═══════════╦═══════════╦═════════╦═══════╗
║ SELLERID ║ SHOP_NAME ║ PRODUCTID ║ PAGEVIEWS ║ TITLE ║ PRICE ║
╠══════════╬═══════════╬═══════════╬═══════════╬═════════╬═══════╣
║ 1 ║ mitienda ║ 2 ║ 30 ║ bufanda ║ $25 ║
║ 3 ║ new_world ║ 6 ║ 6 ║ ropa ║ $13 ║
╚══════════╩═══════════╩═══════════╩═══════════╩═════════╩═══════╝