开发者

MySQL how to do COUNT(*) on a query with joins?

开发者 https://www.devze.com 2023-03-11 16:48 出处:网络
Does anyone know how can I perform COUNT(*) on a query like this? SELECT reviews.review_id, reviews.reviewers_rating, reviews.reviewers_name, reviews.review_date, reviews.pros, reviews.cons, products

Does anyone know how can I perform COUNT(*) on a query like this?

SELECT reviews.review_id, reviews.reviewers_rating, reviews.reviewers_name, reviews.review_date, reviews.pros, reviews.cons, products.product_name, products.slug, products.community_rating, products.number_of_votes, products.users_rating, products.thumb_link, categories.category_name_single
FROM reviews
LEFT JOIN products ON reviews.product_id = products.product_id
LEFT JOIN categories ON products.category = categories.category_id
WHERE reviews.approved 开发者_Python百科=1
ORDER BY reviews.review_date DESC

Thanks!


Try this:

SELECT Count(reviews.review_id), reviews.review_id, reviews.reviewers_rating, reviews.reviewers_name,  reviews.review_date, reviews.pros, reviews.cons, products.product_name, products.slug, products.community_rating, products.number_of_votes, products.users_rating, products.thumb_link, categories.category_name_single
FROM reviews
LEFT JOIN products ON reviews.product_id = products.product_id
LEFT JOIN categories ON products.category = categories.category_id
WHERE reviews.approved =1
Group By reviews.review_id
ORDER BY reviews.review_date DESC


Left Join's are not needed, same as order, and you don't have to count all fields, one will be enough

SELECT COUNT(review_id) FROM reviews WHERE approved =1

EDIT:
if you want to count reviews for products than you should make something like this:

SELECT products.*, count(r.review_id) AS reviews_count FROM products AS p
LEFT JOIN reviews AS r ON p.product_id = r.product_id AND r.approved=1
GROUP BY products.product_id


SELECT COUNT(*) FROM reviews
LEFT JOIN products ON reviews.product_id = products.product_id
LEFT JOIN categories ON products.category = categories.category_id
WHERE reviews.approved = 1 AND ...
ORDER BY reviews.review_date DESC
0

精彩评论

暂无评论...
验证码 换一张
取 消