mysql 查找重复姓名且年龄最大的列表
mysql> select count(*) as count ,name,sum(age) as age from t1 group by name order by count desc;+-------+--------+------+| count | name | age |+-------+--------+------+| 3 | atest | 64 || 2 | btest | 37 || 2 | ctest | 43 || 2 | dtest | 43 || 1 | mary | 22 || 1 | kou | 22 || 1 | perter | 23 || 1 | kate | 19 |+-------+--------+------+8 rows in set (0.00 sec)
这里找到count 重复的数据
下面接着找 count 最大,切age 最大且相同的数据
mysql> select count,name,age from ( select count(*) as count ,name,sum(age) as age from t1 group by name order by count desc ,age desc ) as tmp group by count order by count desc ,age desc;+-------+--------+------+| count | name | age |+-------+--------+------+| 3 | atest | 64 || 2 | ctest | 43 || 1 | perter | 23 |+-------+--------+------+3 rows in set (0.00 sec)
为什么少了一条 dtest ,dtest的数据和ctest在count和age上是一样的?
求指教!谢谢
回复讨论(解决方案)
第二式有 group by count,那么 count 相同的肯定在一组了
既然
| 2 | ctest | 43 |
| 2 | dtest | 43 |
在一组,那自然只能出现一个了
所以分组条件应加上 age,即 group by count,age
mysql> select count(*) as count ,name,sum(age) as age from t1 group by name order by count desc, ae desc ;+-------+--------+------+| count | name | age |+-------+--------+------+| 3 | zx | 64 || 2 | xz | 43 || 2 | john | 43 || 2 | tom | 37 || 1 | perter | 23 || 1 | mary | 22 || 1 | kou | 22 || 1 | kate | 19 |+-------+--------+------+8 rows in set (0.00 sec)mysql> select count,name,age from ( select count(*) as count ,name,sum(age) as age from t1 group b name order by count desc ,age desc ) as tmp group by count,age order by count desc ,age desc;+-------+--------+------+| count | name | age |+-------+--------+------+| 3 | zx | 64 || 2 | xz | 43 || 2 | tom | 37 || 1 | perter | 23 || 1 | mary | 22 || 1 | kate | 19 |+-------+--------+------+6 rows in set (0.00 sec)
这个结果不对吧?
预期的结果应该是:
+-------+--------+------+| count | name | age |+-------+--------+------+| 3 | atest | 64 || 2 | ctest | 43 || 2 | dtest | 43 || 1 | perter | 23 |+-------+--------+------+
select count,name,age from ( select count(*) as count ,name,sum(age) as age from t1 group b
name order by count desc ,age desc ) as tmp group by count, name order by count desc ,age desc;
mysql> select count,name,age from ( select count(*) as count ,name,sum(age) as age from t1 group by name order by count desc ,age desc ) as tmp group by count,name order by count desc ,age desc;+-------+--------+------+| count | name | age |+-------+--------+------+| 3 | zx | 64 || 2 | xz | 43 || 2 | john | 43 || 2 | tom | 37 || 1 | perter | 23 || 1 | kou | 22 || 1 | mary | 22 || 1 | kate | 19 |+-------+--------+------+8 rows in set (0.00 sec)
group by count,name 这样找不到 count=2且age最大的数据和count=1且age最大的数据了~
+-------+--------+------+| count | name | age |+-------+--------+------+| 3 | atest | 64 || 2 | ctest | 43 || 2 | dtest | 43 || 1 | perter | 23 |+-------+--------+------+
这种预期的结果,能在一条sql里体现出来么?
求指教
感觉效率应该不是很高,虽然可以做出来
(select count,max(age) as age from ( select count(*) as count ,name,sum(age) as age from t1 group by name order by count desc ,age desc ) as tmp group by count order by count desc ,age desc) as tempd,( select count(*) as count ,name,sum(age) as age from t1 group by name order by count desc ,age desc ) as tmp1 where tmp1.count=tempd.count and tmp1.age=tempd.age count order by tempd.count desc ,tempd.age desc;
select tmp1.count,tmp1.age from (select count,max(age) as age from ( select count(*) as count ,name,sum(age) as age from t1 group by
name order by count desc ,age desc ) as tmp group by count order by count desc ,age desc) as tempd,( select count(*) as count ,name,sum(age) as age from t1 group by
name order by count desc ,age desc ) as tmp1 where tmp1.count=tempd.count and tmp1.age=tempd.age count order by tmp1.count desc ,tmp1.age desc;
非常感谢版主

热AI工具

Undresser.AI Undress
人工智能驱动的应用程序,用于创建逼真的裸体照片

AI Clothes Remover
用于从照片中去除衣服的在线人工智能工具。

Undress AI Tool
免费脱衣服图片

Clothoff.io
AI脱衣机

AI Hentai Generator
免费生成ai无尽的。

热门文章

热工具

记事本++7.3.1
好用且免费的代码编辑器

SublimeText3汉化版
中文版,非常好用

禅工作室 13.0.1
功能强大的PHP集成开发环境

Dreamweaver CS6
视觉化网页开发工具

SublimeText3 Mac版
神级代码编辑软件(SublimeText3)

热门话题

JWT是一种基于JSON的开放标准,用于在各方之间安全地传输信息,主要用于身份验证和信息交换。1.JWT由Header、Payload和Signature三部分组成。2.JWT的工作原理包括生成JWT、验证JWT和解析Payload三个步骤。3.在PHP中使用JWT进行身份验证时,可以生成和验证JWT,并在高级用法中包含用户角色和权限信息。4.常见错误包括签名验证失败、令牌过期和Payload过大,调试技巧包括使用调试工具和日志记录。5.性能优化和最佳实践包括使用合适的签名算法、合理设置有效期、

文章讨论了PHP 5.3中引入的PHP中的晚期静态结合(LSB),从而允许静态方法的运行时分辨率调用以获得更灵活的继承。 LSB的实用应用和潜在的触摸

使用PHP的cURL库发送JSON数据在PHP开发中,经常需要与外部API进行交互,其中一种常见的方式是使用cURL库发送POST�...

SOLID原则在PHP开发中的应用包括:1.单一职责原则(SRP):每个类只负责一个功能。2.开闭原则(OCP):通过扩展而非修改实现变化。3.里氏替换原则(LSP):子类可替换基类而不影响程序正确性。4.接口隔离原则(ISP):使用细粒度接口避免依赖不使用的方法。5.依赖倒置原则(DIP):高低层次模块都依赖于抽象,通过依赖注入实现。

深入解读ReactPHP的非阻塞特性ReactPHP的一段官方介绍引起了不少开发者的疑问:“ReactPHPisnon-blockingbydefault....
