本文介绍了MYSQL COUNT返回NULL?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我用谷歌搜索了问题,但没有得到答案。
我想列出以下sql的所有结果,包括NULL(当COUNT(review.id)也返回0时),但是我只得到了仅包含评论的地方的结果。
I have googled my problem but didnt get the answer.I want to list all of the results of below sql including NULL (when COUNT(review.id) return 0 also) but instead i just got the results of articles of place that only contains review.
$sql = "SELECT tbl_place.id, tbl_place.region_id, tbl_place.subregion_id, tbl_place.title, tbl_place.metalink, tbl_place.img_thumbnail, tbl_place.summary, tbl_place.category1_id, tbl_place.category2_id, tbl_place.category3_id, COUNT(review.id) AS total_review FROM tbl_place
JOIN review ON tbl_place.id = review.place_id
WHERE
tbl_place.category1_id = '32' AND
tbl_place.status = '1' AND
review.rating != '0.00'
GROUP BY tbl_place.id
ORDER BY total_review $by
LIMIT $limit OFFSET $offset";
推荐答案
请对检查表使用左联接,而不是联接。默认情况下,join是内部联接,因此将只接受匹配的记录。
please use left join for review table instead of join. join is by default inner join so it will take only matched records.
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