双连接mysql查询

我有3个表，但我只能加入另一个表计数。见下文。 下面的一个很有魅力，但我需要从另一个表中添加另一个计数

第三个表名为ci_nomatch，包含对ci_address_book.reference的引用 它可能有多个条目，但我只需要该表的计数

所以如果ci_address_book有一个名为item1，item2，item3的条目 ci_nomatch将有1，item1，user1，2，item1，user4

我想为查询中的Item1返回2

有什么想法吗？我尝试了另一个连接，但它告诉我引用不存在，而它确实存在

SELECT c.*, IFNULL(p.total, 0) AS matchcount
FROM ci_address_book c
LEFT JOIN (
    SELECT addressbook_id, COUNT(match_id) AS total
    FROM ci_matched_sanctions
    GROUP BY addressbook_id
) AS p
ON c.id=p.addressbook_id
ORDER BY matchcount DESC
LIMIT 0,15

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您可以在select中直接子查询它

SELECT c.*, IFNULL(p.total, 0) AS matchcount,
   (SELECT COUNT(*) FROM ci_nomatch n on n.reference = c.reference) AS othercount
FROM ci_address_book c
LEFT JOIN (
    SELECT addressbook_id, COUNT(match_id) AS total
    FROM ci_matched_sanctions
    GROUP BY addressbook_id
) AS p
ON c.id=p.addressbook_id
ORDER BY matchcount DESC
LIMIT 0,15

@更新以供评论。包括一个额外的列matchcount-othercount作为扣除，最好通过子查询来完成

SELECT *, matchcount - othercount AS deducted
FROM
(
    SELECT c.* , IFNULL( p.total, 0 ) AS matchcount, (
        SELECT COUNT( * ) FROM ci_falsepositives n
        WHERE n.addressbook_id = c.reference ) AS othercount
    FROM ci_address_book c
    LEFT JOIN (
        SELECT addressbook_id, COUNT( match_id ) AS total
        FROM ci_matched_sanctions GROUP BY addressbook_id ) AS p
        ON c.id = p.addressbook_id ORDER BY matchcount DESC LIMIT 0 , 15 
) S

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显示第二个连接不起作用的代码，我们将尝试为您调试。[第三个表名为ci_nomatch，包含对ci_address_book.reference的引用]-ci_nomatch中的哪个字段连接到book.reference？我发现它起作用：选择c.*，如果p.total为null，0为matchcount，则从ci_误报n中选择COUNT*n其中n.addressbook_id=c。引用为其他计数从ci_address_book c左加入选择addressbook_id，按c.id=p.addressbook_id按matchcount DESC LIMIT 0，15从ci_matched_制裁组按addressbook_id作为p计算matched总数，是否有可能再产生一个结果，即：matchcount-othercount作为扣除的结果？顺便问一下。。。是的，我知道n.addressbook\u id会让人认为它想删除表addressbook的id字段，但实际上它是addressbook的引用字段，我认为注释被截断了？