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Question

我有一個mysql語句，用於匹配字符串到兩個數據庫中的字符串。我想排除也在另一個表中的結果，但是當我添加conv_progress的左連接時，它會打破語句。任何人有任何想法如何排除在另一個表中的條目？

作品

SELECT video.*, 
    MATCH(video.title) AGAINST(? IN BOOLEAN MODE) as cscore, 
    MATCH(tags.name) AGAINST(? IN BOOLEAN MODE) as htscore 
FROM video 
LEFT JOIN tags ON video.vid_id=tags.vid_id 
null 
WHERE 
    MATCH(video.title) AGAINST(? IN BOOLEAN MODE) OR 
    MATCH(tags.name) AGAINST(? IN BOOLEAN MODE) GROUP BY video.vid_id 
ORDER BY cscore DESC LIMIT ?, ? 

破碎

SELECT video.*, 
    MATCH(video.title) AGAINST(? IN BOOLEAN MODE) as cscore, 
    MATCH(tags.name) AGAINST(? IN BOOLEAN MODE) as htscore 
FROM video 
LEFT JOIN tags ON video.vid_id=tags.vid_id 
LEFT JOIN conv_progress on video.vid_id=conv_progress.vid_id where conv_progress.vid_id is null 
WHERE 
    MATCH(video.title) AGAINST(? IN BOOLEAN MODE) OR 
    MATCH(tags.name) AGAINST(? IN BOOLEAN MODE) GROUP BY video.vid_id 
ORDER BY cscore DESC LIMIT ?, ? 

Answer 1

移動conv_progress.vid_id is null到WHERE子句：

SELECT video.*, 
    MATCH(video.title) AGAINST(? IN BOOLEAN MODE) as cscore, 
    MATCH(tags.name) AGAINST(? IN BOOLEAN MODE) as htscore 
FROM video 
LEFT JOIN tags ON video.vid_id=tags.vid_id 
LEFT JOIN conv_progress on video.vid_id=conv_progress.vid_id 
WHERE 
    (
    MATCH(video.title) AGAINST(? IN BOOLEAN MODE) OR 
    MATCH(tags.name) AGAINST(? IN BOOLEAN MODE) 
) 
    AND conv_progress.vid_id is null 
GROUP BY video.vid_id 
ORDER BY cscore DESC LIMIT ?, ? 

Answer 2

您還可以使用NOT EXISTS而不是加入

SELECT video.*, 
    MATCH(video.title) AGAINST(? IN BOOLEAN MODE) as cscore, 
    MATCH(tags.name) AGAINST(? IN BOOLEAN MODE) as htscore 
FROM video 
LEFT JOIN tags ON video.vid_id=tags.vid_id 
null 
WHERE 
    (
    MATCH(video.title) AGAINST(? IN BOOLEAN MODE) OR 
    MATCH(tags.name) AGAINST(? IN BOOLEAN MODE) 
)AND(NOT EXISTS(SELECT 1 FROM conv_progress WHERE video.vid_id=conv_progress.vid_id)) 
GROUP BY video.vid_id 
ORDER BY cscore DESC LIMIT ?, ?