php - 比较 id 与不同表行 mysql 的 id

Question

所以我试图检索某人的所有兴趣，并能够列出它们。这适用于以下查询。

SELECT *,(
    SELECT GROUP_CONCAT(interest_id SEPARATOR ",")
    FROM people_interests
    WHERE person_id = people.id
) AS interests
FROM people
WHERE id IN (
    SELECT person_id
    FROM people_interests
    WHERE interest_id = '.$site->db->clean($_POST['showinterest_id']).'
)
ORDER BY lastname, firstname

在这个我遇到麻烦的问题中，我只想选择那些碰巧在名为 volleyballplayers 的表中拥有 ID 的人。该表只有 id、person_id、team_id 和日期字段。

SELECT *,(
    SELECT GROUP_CONCAT(interest_id SEPARATOR ",")
    FROM people_interests
    WHERE person_id = people.id
) AS interests
FROM people
WHERE id IN (
    SELECT person_id
    FROM people_interests
    WHERE volleyballplayers.person_id = person_id
)
ORDER BY lastname, firstname

我只是想确保只有 volleyballplayers 表中的人员出现，但我收到一条错误消息，指出 Unknown columns 'volleyballplayers.person_id' in 'where Clause'我非常确定表的名称，并且我知道该列名为 person_id。

Answer 1

尝试将其与子查询连接起来，

SELECT  *, GROUP_CONCAT(interest_id) interests
FROM    people a
        INNER JOIN people_interests b
            ON b.person_id = a.id
        INNER JOIN
        (
            SELECT DISTINCT person_id
            FROM volleyballplayers
        ) c ON b.person_id = c.person_id
GROUP BY a.id
ORDER BY lastname, firstname