Question

我已经用所需的过滤器查询了user_invitations.inviter_id：

select 
  inviter_id, 
  count(inviter_id) as invited_cnt
from user_invitations
group by 
  inviter_id
having 
  invited_cnt >= 52
order by 
  invited_cnt

在这里，我尝试仅使用邀请者ID来获取另一个名为用户的表中的ID（即，要获取与我刚刚获得的邀请者ID相匹配的user_id）

我尝试做内在的调动

from users
inner join (
    select inviter_id, count(inviter_id) as invited_cnt
    from user_invitations
    group by inviter_id
    having invited_cnt >= 52
    order by invited_cnt
)user_inv
on users.id = user_inv.id;

不走运，我考虑过使用子查询

from users
where id in (
    select inviter_id
    from user_invitations
    where inviter_id in (
        select count(inviter_id) as invited_cnt
        from user_invitations
        group by inviter_id
        having invited_cnt >= 52
    )
    group by inviter_id
);

但是在大多数内部查询中，我无法找到一种方法来通过vited_cnt> = 52进行过滤后返回inviter_id。我如何实现这一点，因为使用了inved_cnt> = 52，我们是否需要总计数是不可能的吗？ / p>

我查看了一些资源： Select id from one query and use it in another

但它们似乎都不处理聚合。

Answer 1

如果我理解您的问题，则此查询应为您提供%{ #include <iostream> using namespace std; extern int yylex(); %} %option noyywrap %% . { fwrite(yytext, yyleng, 1, yyout); } %% int main(int argc, char *argv[]) { FILE *sourceFile = fopen("swflow.dfd", "r"); FILE *genFile; if(sourceFile) { genFile = fopen("swflow.v", "w+"); } else { cout << "Cannot open source file!" << endl; return -1; } yyin = sourceFile; yyout = genFile; while(yylex()); return 0; }中与user_id匹配的user

inviter_id

如果您只期望一个SELECT us.inviter_id, count(us.inviter_id) AS invited_cnt, u.user_id FROM user_invitations AS us JOIN user AS u ON u.user_id = us.inviter_id //just change the column name if needed GROUP BY us.inviter_id HAVING invited_cnt >= 52 ORDER BY invited_cnt，那么我认为这应该可行。尽管包括表结构以及一些示例数据和预期输出会很好

从查询中获取ID并在另一个查询中使用它

1 个答案: