MySQL:我需要hel p子查询

尝试以下操作:

-- WITH ORDER BY COUNT(r.reserva_id) DESC LIMIT 1
SELECT  c.client_id, COUNT(r.reserva_id) `count`
FROM reserves r
INNER JOIN clients c ON c.client_id = r.client_id
GROUP BY c.client_id
ORDER BY 2 DESC
LIMIT 1;

-- WITH SUB QUERIES
SELECT B.*
FROM
(SELECT  c.client_id, COUNT(r.reserva_id) `count`
FROM reserves r
INNER JOIN clients c ON c.client_id = r.client_id
GROUP BY c.client_id) B WHERE B.`count`=
(SELECT MAX(A.`count`)
FROM
(SELECT  c.client_id, COUNT(r.reserva_id) `count`
FROM reserves r
INNER JOIN clients c ON c.client_id = r.client_id
GROUP BY c.client_id) A);

看到了吗 run on SQL Fiddle .

对我来说,只需执行第一个查询,它很容易理解和运行。

首先 join 对于您编写的查询不是必需的。

第二,我可以把你的问题理解为“最有保留的客户”。当前查询中的逻辑与此不符。

但对于此版本的查询,它将是:

SELECT  r.client_id, COUNT(*)
FROM reserves r
GROUP BY r.client_id
HAVING COUNT(*) = (SELECT COUNT(*) 
                   FROM reserves r2
                   GROUP BY r2.client_id
                   ORDER BY COUNT(*) DESC
                   LIMIT 1
                  );

如果你真的想避免 ORDER BY / LIMIT 1 在子查询中,可以执行以下操作:

SELECT  r.client_id, COUNT(*)
FROM reserves r
GROUP BY r.client_id
HAVING COUNT(*) = (SELECT MAX(cnt)
                   FROM (SELECT COUNT(*) 
                         FROM reserves r2
                         GROUP BY r2.client_id
                        ) r2
                   );