Question

我有一个数据库，我正在努力进行查询。主表格为people，如果add，{I} id仅在favourites的{{1}}结尾处client_member_id }匹配。

这个想法是循环遍历每个People行，如果那个人ID在f.id表中并且它与{{favourites匹配，则在id的行末尾添加client_member_id 1 {} client_member_id。

1是给每个客户的ID，这样当他们最喜欢一个人时，细节就会被放到收藏夹表中。这可以是2或FAVE_ID。在SQLFiddle示例中，client_member_id为2时68的预期输出为69 client_member_id，但如果1为FAVE_ID然后null都是SELECT *, j.company_id as companyid, j.id as jid, p.id as pid, (CASE WHEN f.client_member_id = 2 AND p.id = f.people_id THEN f.id ELSE null END) AS fave_id, (CASE WHEN f.client_member_id = 2 AND p.id = f.people_id THEN 1 ELSE 0 END) AS fave FROM people p INNER JOIN favourites f INNER JOIN job j ON p.job_id = j.id WHERE p.company_id = 1 ORDER BY p.id ASC

我已经把它弄到了似乎复制行的地步，我需要它是截然不同的。

{{1}}

SQLFiddle

提前谢谢。

Answer 1

没有重复项，因为fave_id和fave列，你有4行。如果你想过滤它们，那么这里是一个如何只显示fave = 1的值的例子：

SELECT *, j.company_id as companyid, j.id as jid, p.id as pid,
    (CASE WHEN f.client_member_id = 2 AND p.id = f.people_id THEN f.id ELSE null END) AS fave_id,
    (CASE WHEN f.client_member_id = 2 AND p.id = f.people_id THEN 1 ELSE 0 END) AS fave
FROM people p
INNER JOIN favourites f
INNER JOIN job j 
ON p.job_id = j.id
WHERE p.company_id = 1 AND f.client_member_id = 2 AND p.id = f.people_id
ORDER BY p.id ASC

Answer 2

您可以使用LEFT JOIN加入收藏夹以摆脱笨拙的CASE：

    SELECT *,
           j.company_id companyid,
           j.id jid,
           p.id pid,
           f.id fave_id,
           f.id IS NOT NULL fave
      FROM people p    
INNER JOIN job j 
        ON p.job_id = j.id
 LEFT JOIN favourites f
        ON f.people_id=p.id
       AND f.client_member_id=2
     WHERE p.company_id = 1
  ORDER BY p.id ASC

NB 这将为您的小提琴只返回两行fave_id和fave，NOT NULL和1，但如果您有一个没有相应的人最喜欢的是你得到fave_id和fave，NULL和0。

Answer 3

试试这个SQLFiddle

SELECT *, j.company_id as companyid, j.id as jid, p.id as pid,
    (CASE WHEN f.client_member_id = 2 THEN f.id ELSE null END) AS fave_id,
    (CASE WHEN f.client_member_id = 2 THEN 1 ELSE 0 END) AS fave
FROM people p
INNER JOIN job j ON p.job_id = j.id
left JOIN favourites f on f.people_id = p.id
WHERE p.company_id = 1
ORDER BY p.id ASC

您没有将您的收藏夹表连接绑定到任何内容，因此您有多少行*返回的收藏数。我使用左连接，因为我猜你不能保证人员表中的每个人都会有一个喜欢的东西？否则，如果你可以改为内连接。

此致

利安

MySQL高级选择查询检查来自另一个表的数据

3 个答案: