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MYSQL COUNT返回NULL?

[英]MYSQL COUNT return NULL?

I have googled my problem but didnt get the answer. 我用谷歌搜索了我的问题,但没有得到答案。 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. 我想列出以下所有包含NULL的sql的结果(当COUNT(review.id)也返回0时),但是我只是得到了仅包含评论的地方的结果。

$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";

please use left join for review table instead of join. 请使用左联接而不是联接查看表。 join is by default inner join so it will take only matched records. 缺省情况下,join是内部联接,因此它将仅接受匹配的记录。

the sql should be : sql应该是:

$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, 
(SELECT COUNT(*) FROM review WHERE review.rating != '0.00' AND tbl_place.id = review.place_id ) AS total_review 
FROM tbl_place WHERE 
tbl_place.category1_id = '32' AND 
tbl_place.status = '1'   
GROUP BY tbl_place.id 
ORDER BY total_review $by";

it's working! 工作正常! thx guys! 谢谢你们!

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