MySQL：我需要帮助Subquery

Posted 2021-04-09

嗨，我对这个子查询有一个问题，如果可能的话，我想知道客户对他们的名字有更多的保留而不使用ORDER BY .. DESC LIMIT 1;

我有这个代码：

SELECT  c.client_id, COUNT(r.reserva_id)
    FROM reserves r
INNER JOIN clients c ON c.client_id = r.client_id
GROUP BY c.client_id
HAVING COUNT(r.reserva_id) < (SELECT COUNT(r2.reserva_id) 
                        FROM reserves r2
                     GROUP BY r2.client_id)

但它不起作用。你能帮助我吗

谢谢..

答案

试试这个：

-- 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);

看到它run on SQL Fiddle。

对我来说，只需使用第一个查询，它很容易理解和运行。

另一答案

首先，对于您编写的查询，join不是必需的。

其次，我可以将您的问题解释为“保留最多的客户”。您当前查询中的逻辑与此无关。

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
                  );

如果你真的想在子查询中避免使用ORDER BY / LIMIT 1，你可以这样做：

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
                   );