Question

现在我有三个mysql表。我必须在表格中列出people中的所有人。接下来它显示了兴趣，这里是它的sql。

SELECT *,(
           SELECT GROUP_CONCAT(interest_id SEPARATOR ",")
           FROM cat_people_interests
           WHERE person_id = people.id
         ) AS cat_interests
FROM people
WHERE id IN 
(
    SELECT person_id
    FROM cat_interests
)
ORDER BY lastname, firstname);

SCHEMA

`people` { int:id, varchar:name }
`cat_people_interests` { int:interest_id, int:person_id }
`cat_interests` { int:id, varchar:name }
`alternate_meow_pull` { int:person_id, int:meow_id, int:dataetc, int:dataetc2 }

所以这让我有能力把名字和所有兴趣都拉到那里。

如果一个人的id在另一个表（alternate_meow_pull）中作为person_id存在，我如何更改或添加到每行来检查？我个人所做的只是检查每一行的mysql查询，但我确信有更快更简洁的方法。我想得到结果，无论这个人是否存在于alternate_meow_pull中。

因此，当我遍历此结果时，我希望能够打印名称，ID，与人相关的所有兴趣，以及恰好位于与人相关的alternate_meow_pull中的任何数据。

Answer 1

不应该是这样吗？

SELECT  a.name, 
        GROUP_CONCAT(b.name) InterestList
FROM    people  a
        INNER JOIN cat_people_interest c
            ON a.ID = c.person_ID
        INNER JOIN cat_interests b
            ON b.id = c.interest_ID 
        INNER JOIN alternate_meow_pull d
            ON a.person_ID = d.person_id
WHERE   b.interest_id = '$_POST['showinterest_id']'
GROUP BY a.name

只会显示ID上alternate_meow_pull出现的人的姓名。

Answer 2

 SELECT people.id, people.name
 FROM people,cat_people_interests as c,cat_interests as i, alternate_meow_pull as m
 WHERE people.id = c.person_id or people.id = i.id or people.id = m.person_id

如果行的id作为另一个表中的id存在，则每行检查一次

2 个答案: