Question

嗨我这个子查询有问题我想知道如果可能的话，如果可能的话，在不使用ORDER BY的情况下对客户有更多保留的客户端.DESC 限制1;

我有这段代码：

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。你能帮我吗

谢谢..

Answer 1

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);

见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。

Answer 2

首先，您编写的查询不需要join。

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
                   );

MySQL：我需要hel p子查询

2 个答案: