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我的應用程序利用關係表的表格,我用我的SQL連接視頻與類別,演員和標籤GROUP_CONCAT與DISTINCT。獨特不消除空結果
我面臨的問題是我會爲演員輸出拇指,但有時候演員還沒有縮略圖。在這種情況下,我在group_concat的行中產生的值與actor字段不一致,因爲3個actor是唯一的,但是如果我有2個actor沒有圖像(null),那麼我失去了錨點。
例如
SELECT
videos.id AS id,
videos.video_title AS video_title,
videos.video_views AS video_views,
videos.video_likes AS video_likes,
videos.video_dislikes AS video_dislikes,
videos.video_duration AS video_duration,
GROUP_CONCAT(DISTINCT a.actor_name SEPARATOR ';') AS actor_names,
GROUP_CONCAT(DISTINCT t.tag_name SEPARATOR ';') AS tag_names,
GROUP_CONCAT(DISTINCT c.category_name SEPARATOR ';') AS category_names,
GROUP_CONCAT(DISTINCT a.actor_thumb SEPARATOR ';') AS actor_thumbs
FROM videos
LEFT OUTER JOIN video_actors AS va ON va.video_id = videos.id
LEFT OUTER JOIN actor AS a ON a.actor_id = va.actor_id
LEFT OUTER JOIN video_tags AS vt ON vt.video_id = videos.id
LEFT OUTER JOIN tags AS t ON t.tag_id = vt.tag_id
LEFT OUTER JOIN video_categories AS vc ON vc.video_id = videos.id
LEFT OUTER JOIN categories AS c ON c.category_id = vc.category_id
WHERE videos.id = '23'
演員拇指場結果:
more fields.. |0;http://site.com/actor/59.jpg
但它應該是(3演員,2沒有縮略圖):
more fields.. |0;0;http://site.com/actor/59.jpg
爲了保持對演員的名字值一致。
希望這有點清楚。
提前致謝! 尼克
編輯答案,actully我的第一個建議直接難道不具有鮮明的工作。 – barryhunter
感謝您的解決方案 – directory