MYSQL COUNT返回NULL？

我GOOGLE了我的問題，但沒有得到答案。我想列出下面的SQL的所有結果，包括NULL（當COUNT（review.id）也返回0），而是我只是得到了只包含評論的地方文章的結果。MYSQL COUNT返回NULL？

$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";

來源

2012-03-20 oit oit

可能重複[Mysql：Count記錄（包括零）每月]（http://stackoverflow.com/questions/3486998/mysql-count-records-including-zero-per-month）。有很多，如果你看看：http://stackoverflow.com/search?q=[mysql]+count+zero – 2012-03-20 04:12:39

thx OMG小馬，我看了一下，有一個答案！ – 2012-03-20 04:37:40

請使用左連接作爲審查表而不是連接。連接默認情況下是內部連接，所以它將只接受匹配的記錄。

來源

2012-03-20 04:29:55

我找到了答案！ – 2012-03-20 04:33:02

thx btw Vetrivel mp！ – 2012-03-20 04:33:24

的sql應該是：

$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";

它的工作！ thx傢伙！

來源

2012-03-21 03:45:34

MYSQL COUNT返回NULL？

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