Question

我有3张桌子，但我只能加入另一张桌子。见下文。下面的一个就像一个魅力，但我需要从另一个表中添加另一个“计数”。

有一个名为“ci_nomatch”的第3个表，其中包含对ci_address_book.reference的引用可能有多个条目（许多条目很多），但我只需要该表的计数。

所以如果ci_address_book有一个名为“item1”，“item 2”，“item3”的条目和ci_nomatch会有“1，item1，user1”，“2，item1，user4”

我想在查询中为Item1返回“2”。

有什么想法吗？我尝试了另一个连接，但它告诉我该引用不存在，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

Answer 1

您可以直接在选择

中对其进行子查询

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

@updated发表评论。包括额外的列“（matchcount - othercount）AS扣除”最好通过查询来完成。

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

带双连接的mysql查询

1 个答案: