Mysql根据指定字段的int值查出在当前列表的排名

Posted 2020-09-18 路漫漫其修远兮

先看表结构和数据：

DROP TABLE IF EXISTS `ndb_record`;
CREATE TABLE `ndb_record` (
  `id` bigint(20) NOT NULL AUTO_INCREMENT COMMENT \'测量记录\',
  `user_id` bigint(20) NOT NULL COMMENT \'用户id\',
  `yellow` int(11) DEFAULT NULL COMMENT \'黄色状态持续时长\',
  `green` int(11) DEFAULT NULL COMMENT \'绿色状态持续时长\',
  `blue` int(11) DEFAULT NULL COMMENT \'蓝色状态时长\',
  `create_time` date DEFAULT NULL COMMENT \'测量时间\',
  `week` varchar(20) DEFAULT NULL COMMENT \'周几\',
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=35 DEFAULT CHARSET=utf8;

-- ----------------------------
-- Records of ndb_record
-- ----------------------------
INSERT INTO `ndb_record` VALUES (\'17\', \'13\', \'8\', \'7\', \'6\', \'2017-03-23\', \'星期四\');
INSERT INTO `ndb_record` VALUES (\'18\', \'13\', \'8\', \'7\', \'6\', \'2017-03-22\', \'星期三\');
INSERT INTO `ndb_record` VALUES (\'19\', \'13\', \'8\', \'7\', \'6\', \'2017-03-20\', \'星期一\');
INSERT INTO `ndb_record` VALUES (\'20\', \'13\', \'8\', \'7\', \'6\', \'2017-03-19\', \'星期日\');
INSERT INTO `ndb_record` VALUES (\'21\', \'13\', \'8\', \'7\', \'6\', \'2017-03-18\', \'星期六\');
INSERT INTO `ndb_record` VALUES (\'22\', \'13\', \'8\', \'7\', \'8\', \'2017-03-23\', \'星期四\');
INSERT INTO `ndb_record` VALUES (\'23\', \'13\', \'8\', \'7\', \'1\', \'2017-03-20\', \'星期一\');
INSERT INTO `ndb_record` VALUES (\'24\', \'13\', \'8\', \'7\', \'2\', \'2017-03-14\', \'星期二\');
INSERT INTO `ndb_record` VALUES (\'25\', \'13\', \'8\', \'7\', \'3\', \'2017-03-17\', \'星期五\');
INSERT INTO `ndb_record` VALUES (\'26\', \'13\', \'8\', \'7\', \'4\', \'2017-03-16\', \'星期四\');
INSERT INTO `ndb_record` VALUES (\'27\', \'12\', \'8\', \'7\', \'4\', \'2017-03-21\', \'星期二\');
INSERT INTO `ndb_record` VALUES (\'28\', \'12\', \'8\', \'7\', \'4\', \'2017-03-20\', \'星期一\');
INSERT INTO `ndb_record` VALUES (\'29\', \'12\', \'8\', \'7\', \'4\', \'2017-03-20\', \'星期一\');
INSERT INTO `ndb_record` VALUES (\'30\', \'12\', \'6\', \'7\', \'4\', \'2017-03-19\', \'星期日\');
INSERT INTO `ndb_record` VALUES (\'31\', \'12\', \'6\', \'7\', \'3\', \'2017-03-18\', \'星期六\');
INSERT INTO `ndb_record` VALUES (\'32\', \'16\', \'6\', \'7\', \'3\', \'2017-03-16\', \'周四\');
INSERT INTO `ndb_record` VALUES (\'33\', \'16\', \'6\', \'7\', \'3\', \'2017-03-31\', \'周五\');
INSERT INTO `ndb_record` VALUES (\'34\', \'16\', \'6\', \'6\', \'0\', \'2017-04-05\', \'周三\');

她给出的问题是，通过这条Sql语句统计了每个字段的总和，然后找出指定user_id关联times总和的排名

SELECT user_id,(SUM(yellow)+SUM(green)+SUM(blue)) AS times FROM ndb_record GROUP BY user_id;

查询出的结果是：

我给出了两种方法一条SQL实现。

第一种

 SELECT o_d FROM (SELECT a.*, 
       @rownum := @rownum + 1 AS o_d
  FROM (
SELECT user_id,(SUM(yellow)+SUM(green)+SUM(blue)) AS times FROM ndb_record GROUP BY user_id ORDER BY times DESC
) a, 
       (SELECT @rownum := 0) r) b WHERE user_id =13

第二种

SELECT
    count(*) AS o_d
FROM
    (
        SELECT
            user_id,
            (
                SUM(yellow) + SUM(green) + SUM(blue)
            ) AS times
        FROM
            ndb_record
        GROUP BY
            user_id
    ) a
WHERE
    times >= (
        SELECT
            times
        FROM
            (
                SELECT
                    user_id,
                    (
                        SUM(yellow) + SUM(green) + SUM(blue)
                    ) AS times
                FROM
                    ndb_record
                GROUP BY
                    user_id
            ) b
        WHERE
            `user_id` = 13
    )

查询结果也是跟第一种一样。

 

可能以上说明您没太明白，然后我再拿一条简单的表举例：

CREATE TABLE `test` (
  `id` int(11) NOT NULL AUTO_INCREMENT COMMENT \'主键\',
  `name` varchar(22) NOT NULL DEFAULT \'\' COMMENT \'姓名\',
  `age` int(11) NOT NULL DEFAULT \'0\' COMMENT \'年龄\',
  PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=6 DEFAULT CHARSET=utf8

表创建好了，看后查看一下结果SELECT * FROM test：

比如，我们要查的是王五在这五个人里年龄排第几，目测赵六是老大，也就是排名第一，刘七老五，排名第五。

上语句：

select * from (SELECT t.*, 
       @rownum := @rownum + 1 AS o_d
  FROM (
select * from test order by age desc
) t, 
       (SELECT @rownum := 0) r) b where id =1

查询的条件是id=1，也就是张三，结果是4.

SELECT count(*) AS o_d FROM (SELECT age FROM test) a WHERE age >= (SELECT age FROM (SELECT * FROM test)
 b WHERE `id`=\'1\');

完毕。

 

虽然结果出来了，还请前辈们多多指教哪里的不足！致敬！~