mysql多表连接和计数查询

我总共有6个表，其中保存了不同的信息

现在我需要一个从5个表中获取计数并从主表中选择所有信息的结果，但如果记录不存在，则必须返回0，而不是找不到行，这就是问题所在

我尝试了下面的查询，但没有成功

SELECT 
    u.*,
    COUNT(DISTINCT  c.id) as comments,
    COUNT(DISTINCT d.id) as dislikes,
    COUNT(DISTINCT l.id) as likes,
    COUNT(DISTINCT s.id) as shares,
    COUNT(DISTINCT t.id) as tags
FROM 
    job_details as u
    JOIN job_comments as c ON u.id = c.job_id
    JOIN job_dislike as d ON u.id = d.job_id
    JOIN job_like as l ON u.id = l.job_id
    JOIN job_share as s ON u.id = s.job_id
    JOIN job_tags as t ON u.id = t.job_id
WHERE 
    u.id = c.job_id AND
    u.id = d.job_id AND
    u.id = l.job_id AND
    u.id = s.job_id AND
    u.id = t.job_id

GROUP BY 
    u.id

已执行此查询，但未得到确切结果。 我不太明白为什么

我希望这里有人能帮我

---

谢谢

使用

左连接

而不是

连接

。而且您不需要

WHERE

子句，因为您已经加入了这些表。并且，使用

IFNULL

函数为空值返回0。您需要修改查询，如下所示：

SELECT u.id,
    IFNULL(COUNT(DISTINCT  c.id),0) as comments,
    IFNULL(COUNT(DISTINCT d.id),0) as dislikes,
    IFNULL(COUNT(DISTINCT l.id),0) as likes,
    IFNULL(COUNT(DISTINCT s.id),0) as shares,
    IFNULL(COUNT(DISTINCT t.id),0) as tags
FROM job_details as u
    LEFT JOIN job_comments as c ON u.id = c.job_id
    LEFT JOIN job_dislike as d ON u.id = d.job_id
    LEFT JOIN job_like as l ON u.id = l.job_id
    LEFT JOIN job_share as s ON u.id = s.job_id
    LEFT JOIN job_tags as t ON u.id = t.job_id
GROUP BY u.id

---

您可能没有得到确切的结果，因为某些表可能缺少值

虽然您可以通过

左连接解决此问题，但更安全的解决方案是预聚合数据：

SELECT u.*, c.comments, d.dislikes, l.likes, s.shares, t.tags
FROM job_details as u LEFT JOIN
     (select c.job_id, count(*) as comments from job_comments group by c.job_id
     ) c
     ON u.id = c.job_id LEFT JOIN
     (select d.job_id, count(*) as dislikes from job_dislike d group by d.job_id
     ) d
     ON u.id = d.job_id LEFT JOIN
     (select l.job_id, count(*) as likes from job_like l group by l.job_id
     ) l
     ON u.id = l.job_id LEFT JOIN
     (select s.job_id, count(*) as shares from job_share s group by s.job_id
     ) s
     ON u.id = s.job_id LEFT JOIN
     (select t.job_id, count(*) as tags from job_tags t group by t.job_id
     ) t
     ON u.id = t.job_id;

为什么这样更好？考虑一个ID有5个评论，喜欢，不喜欢，股票和标签。
JOIN
方法生成中间结果，中间行数为5*5*5*5*5=3125。对于流行的ID，情况可能真的会失控。
发布一些示例数据、实际结果和预期结果。由于存在内部联接，您可能会获得比预期更少的数据…您应该检查哪个是主表，并使用左联接联接其他表。是否可以在此查询中添加where条件？@AnkitDoshi。对在外部查询中或在

on子句中，具体取决于您希望发生什么。