在销售方面找到最优秀的员工。 [英] Find the top employees in terms of sales.

问题描述

需要在销售方面找到最优秀的员工，但我无法获得足够的产出



我尝试过：

在e.e_id = s.e_id group中按员工e加盟销售选择e_name by sale_id limit 1; 

解决方案

您需要对查询进行排序以将最高值带到顶部 - 提示：按降序排列。



参考： MySQL :: MySQL 8.0参考手册:: 8.2.1.14 ORDER BY优化 [ ^ ]

只是添加到CHill60所说的内容：您需要ORDER BY销售，但是因为您正在使用GROUP BY，你还需要查看一个合适的聚合函数。



看看这里： [ ^ ]因为这是咬你的下一件事 - 它是基于SQL Server但是MySql的原理是相同的

in need to find the top employee in terms of sales but i am not able to get sufficient output

What I have tried:

select e_name from employee e join sale s on e.e_id=s.e_id group by sale_id limit 1;

解决方案

You need to ORDER your query to bring the top value to the "top" - hint: in descending order.

Reference: MySQL :: MySQL 8.0 Reference Manual :: 8.2.1.14 ORDER BY Optimization[^]

Just to add to what CHill60 says: you need to ORDER BY sales, but since you are using GROUP BY, you will also need to look at an appropriate aggregate function.

Have a look here: SQL GROUP By and the "Column 'name' is invalid in the select list because..." error[^] as that's the next thing that will bite you - it's SQL Server based but the principles are the same for MySql

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