Question

我有一个复杂的INNER JOIN SQL请求我无法解决这个问题。我希望有人可以帮助我。它涉及3个表，因此它有2个内连接。

我的数据库有“用户”，“声明”，“意见”等表格。

用户可以撰写陈述和意见。意见有一个“authorid”变量引用它们所代表的用户的id，并且它们有一个“statementid”变量引用它们所引用的语句。

我正在尝试提交请求，在给出2个语句的情况下，我可以返回已撰写关于这两个语句的意见的用户列表。

我在想像

$sid1=5;
$sid2=6;
$sql = "
        SELECT user.*
        FROM users
        INNER JOIN opinion
        ON opinion.authorid=user.uid 
        WHERE opinion.statementid=  [sid1?  how can i use both]
        INNER JOIN statement
        ON statment.uid=opinion.statementid
        ";

但是你可以看到我被卡住了。我需要一个UNION吗？如果您需要进一步澄清，请与我们联系。

提前致谢。

编辑：我想出了怎么做：

SELECT DISTINCT users.uid
FROM users
JOIN opinion o, opinion o2
WHERE users.uid = o.authorid
AND users.uid = o2.authorid
AND o2.statementid = $sid2
AND o.statementid = $sid1

Answer 1

  SELECT DISTINCT user.*
  FROM opinion o 
  JOIN usrs u 
       ON u.uid = o.author_id
      AND o.statement_id = $sid1
  INTERSECT
  SELECT DISTINCT user.*
  FROM opinion o 
  JOIN usrs u 
       ON u.uid = o.author_id
      AND o.statement_id = $sid2

您的SQL方言中没有可用的INTERSECT吗？

SELECT A.* FROM (
  SELECT DISTINCT user.*
  FROM opinion o 
  JOIN usrs u 
       ON u.uid = o.author_id
      AND o.statement_id = $sid1
) A
JOIN (
  SELECT DISTINCT user.*
  FROM opinion o 
  JOIN usrs u 
       ON u.uid = o.author_id
      AND o.statement_id = $sid2
) B on A.uid = B.uid

Answer 2

你应该可以连接两次。

SELECT user.*
FROM users u. opinion o1, statement s1
     , opinion o2, statement s2 
WHERE o1.authorid=user.uid 
AND o2.authorid = user.uid
AND o1.statement_id = s1.sid
AND o2.statementid = s2.sid
AND s1.sid = $sid1
AND s2.sid = $sid2

Answer 3

我认为这是您想要的查询：

   SELECT u.*
   from opinion o join
        users u
        on o.authorid = u.uid join
        statement s
        on o.statementid = s.sid
   where s.sid in (<list of statement ids>);

您对数据结构的描述有点令人困惑。但我猜测opinion具有作者和陈述的关键关系（并且statement没有与之关联的单独用户ID）。

Answer 4

SELECT DISTINCT users.uid
FROM users
JOIN opinion o, opinion o2
WHERE users.uid = o.authorid
AND users.uid = o2.authorid
AND o2.statementid = $sid2
AND o.statementid = $sid1

复杂的INNER JOIN SQL请求

4 个答案: