Question

我试图通过聚合函数获取排名值 - 我想看看它是否可以在MySQL查询中完成（而不是ORM调用查询）。

我拥有的表和数据类似于：

/bin/ls

到目前为止，我有：

CREATE TABLE `interactions` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT,
  `account` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
  `timestamp` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM;

INSERT INTO `interactions` VALUES
(1, "xyz", "2017-01-01 00:05:01"),
(2, "xyz", "2017-01-01 00:05:10"),
(3, "abc", "2017-01-01 00:05:21"),
(4, "xyz", "2017-01-01 00:05:43"),
(5, "def", "2017-01-01 00:05:47"),
(6, "xyz", "2017-01-01 00:05:49"),
(7, "abc", "2017-01-01 00:05:50"),
(8, "abc", "2017-01-01 00:05:59");

但是这会输出每个帐户（具有正确的set @curRank := 0; select @curRank := @curRank + 1 AS rank, der.account, der.searches from interactions right join (select account, count(id) AS searches from interactions group by account order by searches) AS der on der.account = interactions.account;值 - 但排名不止一次）：

searches

我正在寻找：

 1    abc        3
 2    abc        3
 3    abc        3
 4    def        1
 5    xyz        4
 6    xyz        4
 7    xyz        4
 8    xyz        4

我应该提一下，我不关心联合排名 - 如果两个账户在表格中以相同的数量结束，那么如果他们一个接一个地排名（或者什么是什么）并不重要顺序）。

Answer 1

向SELECT添加DISTINCT会修复查询：

set @curRank := 0;

select distinct 
    der.account,
    der.searches,
    @curRank := @curRank + 1 AS rank
from
    interactions
right join
    (select account, count(id) AS searches from interactions group by account order by searches) AS der
on
    der.account = interactions.account;

按聚合（计数）排序查询结果

1 个答案: