3個表的mysql查詢

此查詢不選擇，無評論或無表決數據，我如何解決？3個表的mysql查詢

SELECT 
    blog_posts.*, 
    COUNT(DISTINCT(blog_comments.CommentID)) AS CountComments, 
    COUNT(DISTINCT(blog_votes.VoteID)) AS CountVotes, 
    AVG(DISTINCT(blog_votes.Vote)) AS AvgVote 
    FROM blog_posts 
    INNER JOIN blog_comments ON blog_comments.CommentBlogID=blog_posts.BlogID 
    INNER JOIN blog_votes ON blog_votes.VoteBlogID=blog_posts.BlogID 
    ORDER BY blog_posts.BlogID 
    LIMIT 0,10

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2015-09-22 Eren Özkul

'GROUP BY blog_post.col1，blog_posts.col_2，...' – lad2025

所以你想計數評論，投票和平均投票數據？ –

@MichaelAntonio它的真實 –

這個怎麼樣，我使用子查詢，使之簡單：

SELECT b.*, 
    COUNT(select CommentID from blog_comments where CommentBlogID = b.BlogID group by CommentID) as CountComments, 
    COUNT(select VoteID from blog_comments where VoteBlogID = b.BlogID group by VoteID) as CountVotes, 
    AVG(select Vote from blog_comments where VoteBlogID = b.BlogID group by Vote) as AvgVote, 
FROM blog_posts as b 
ORDER BY b.BlogID 
LIMIT 0,10

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2015-09-22 00:58:31

＃1064 - 你的SQL語法有錯誤;請查看與您的MySQL服務器版本相對應的手冊，以便在'SELECT blog_comments.CommentID FROM blog_comments WHERE blog_com'的第3行 –

我固定

SELECT 
blog_posts.*, 
COUNT(VoteID) as CountVotes, 
AVG(VoteID) As AvgVotes, 
COUNT(CommentID) AS CountComments 
FROM blog_posts 
LEFT JOIN blog_votes ON blog_posts.BlogID = blog_votes.VoteBlogID 
LEFT JOIN blog_comments ON blog_posts.BlogID = blog_comments.CommentBlogID 
GROUP BY blog_posts.BlogID 
ORDER BY BlogID DESC 
LIMIT 0,10

來源

2015-09-22 02:16:09

3個表的mysql查詢

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