[英]MYSQL COUNT return NULL?
我用谷歌搜索了我的問題,但沒有得到答案。 我想列出以下所有包含NULL的sql的結果(當COUNT(review.id)也返回0時),但是我只是得到了僅包含評論的地方的結果。
$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";
請使用左聯接而不是聯接查看表。 缺省情況下,join是內部聯接,因此它將僅接受匹配的記錄。
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";
工作正常! 謝謝你們!
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