不要認為GROUP BY是必需的 . 雖然如果一個孩子有2個父記錄，你可能想用它來將2個父母分組到一行 - 但不確定這是否是你的要求 . 因為如果一個孩子有2個父母，那么將為該孩子返回的父母是未定義的 .

假設所有孩子都有父母，所有父母都會有姓，姓和電子郵件： -

SELECT

k.*,

p.*,

u1.meta_value AS parent_lastname,

u2.meta_value AS parent_firstname,

u3.meta_value AS parent_email

FROM kids k

INNER JOIN relations r ON k.ID = r.kid_id

INNER JOIN parents p ON p.user_id = r.user_id

INNER JOIN usermeta u1 ON r.user_id = u1.user_id AND u1.meta_key = 'last_name'

INNER JOIN usermeta u2 ON r.user_id = u1.user_id AND u1.meta_key = 'first_name'

INNER JOIN usermeta u3 ON r.user_id = u1.user_id AND u1.meta_key = 'email'

如果該假設是錯誤的，則將INNER JOIN更改為LEFT OUTER JOIN

SELECT

k.*,

p.*,

u1.meta_value AS parent_lastname,

u2.meta_value AS parent_firstname,

u3.meta_value AS parent_email

FROM kids k

LEFT OUTER JOIN relations r ON k.ID = r.kid_id

LEFT OUTER JOIN parents p ON p.user_id = r.user_id

LEFT OUTER JOIN usermeta u1 ON r.user_id = u1.user_id AND u1.meta_key = 'last_name'

LEFT OUTER JOIN usermeta u2 ON r.user_id = u1.user_id AND u1.meta_key = 'first_name'

LEFT OUTER JOIN usermeta u3 ON r.user_id = u1.user_id AND u1.meta_key = 'email'