mysql 只有最大的计数，有几个最大值答案

【问题标题】：mysql only max of count with several maxmysql 只有最大的计数，有几个最大值
【发布时间】：2018-08-06 19:12:19
【问题描述】：

我正在寻找一个查询，只得到最大的。

SELECT results.playerid, 
       username, 
       Count(*) AS count 
FROM   results 
       LEFT JOIN players 
              ON results.playerid = players.playerid 
WHERE  points = 5 
GROUP  BY playerid 
ORDER  BY count DESC

结果：

+-----------+-----------+-------+
| playerID  | username  | count |
+-----------+-----------+-------+
| 1         | Lorem     | 2     |
| 2         | Ipsum     | 2     |
| 3         | Dolor     | 1     |
+-----------+-----------+-------+

预期输出：

+-----------+-----------+-------+
| playerID  | username  | count |
+-----------+-----------+-------+
| 1         | Lorem     | 2     |
| 2         | Ipsum     | 2     |
+-----------+-----------+-------+

如果只有一个最大的就没有问题了。

有什么建议吗？

小提琴：http://sqlfiddle.com/#!9/7c74c17/1

【问题讨论】：

向我们展示当前和预期的输出。如果是平局，您想退回什么？
顺便说一句，与您的问题无关。但是您应该将COUNT(*) 更改为COUNT(players.playerID) 否则当左连接找不到匹配项时您将不会得到结果0。
谢谢。添加了预期的输出。
你的mysql是什么版本的？
5.6.36-82.0-log

标签： mysql database join count max

【解决方案1】：

首先您需要找出group by 的最大值是多少

SELECT Count(players.playerid) AS count 
FROM   results 
       LEFT JOIN players 
              ON results.playerid = players.playerid 
WHERE  points = 5
GROUP  BY players.playerid
ORDER  BY count DESC
LIMIT 1;

然后您使用该查询来过滤您需要的结果。

SQL DEMO

SELECT results.playerid, 
       username, 
       Count(players.playerid ) AS count 
FROM   results 
       LEFT JOIN players 
              ON results.playerid = players.playerid 
WHERE  points = 5 
GROUP  BY playerid 
HAVING COUNT(players.playerid) = (
          SELECT Count(players.playerid) AS count 
          FROM   results 
                 LEFT JOIN players 
                        ON results.playerid = players.playerid 
          WHERE  points = 5
          GROUP  BY players.playerid
          ORDER  BY count DESC
          LIMIT 1
);

输出

| playerid | username | count |
|----------|----------|-------|
|        1 |    Lorem |     2 |
|        2 |    Ipsum |     2 |

【讨论】：