与SQL MIN()GROUP BY额外的字段
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05-09-2019 - |
题
当使用SQL MIN()函数,与GROUP BY一起将任何附加列(不是MIN列或GROUP BY列之一)的匹配MIN行中的数据相匹配?
例如,与给定的部门名称,雇员姓名,工资和一个表:
SELECT MIN(e.salary), e.* FROM employee e GROUP BY department
很显然,我会得到两个好栏目,最低工资和部门。将员工姓名(和任何其他雇员字段)是由同一行?即与MIN(工资)行?
我知道有可能很可能是两个员工具有相同(最低)的薪水,但我很担心(现在)越来越对所有的信息(或一个的)最便宜的雇员。
这会选择最便宜的推销员?
SELECT min(salary), e.* FROM employee e WHERE department = 'sales'
基本上,可以我肯定的是,数据与所述MIN()函数返回沿将与该最小值的(或单)记录相匹配?
如果数据库的问题,我使用MySQL。
解决方案
如果你想获得各部门中“最便宜”的员工你将有两个选择了我的头顶部:
SELECT
E.* -- Don't actually use *, list out all of your columns
FROM
Employees E
INNER JOIN
(
SELECT
department,
MIN(salary) AS min_salary
FROM
Employees
GROUP BY
department
) AS SQ ON
SQ.department = E.department AND
SQ.min_salary = E.salary
或者你也可以使用:
SELECT
E.*
FROM
Employees E1
LEFT OUTER JOIN Employees E2 ON
E2.department = E1.department AND
E2.salary < E1.salary
WHERE
E2.employee_id IS NULL -- You can use any NOT NULL column here
第二条语句的工作方式是有效的话,告诉我所有的员工,你不能在同一个部门以较低的薪水找到另一名雇员。
在这两种情况下,如果两个或更多的员工有平等的工资是最低,你会得到他们两个(全部)。
其他提示
SELECT e.*
FROM employee e
WHERE e.id =
(
SELECT id
FROM employee ei
WHERE ei.department = 'sales'
ORDER BY
e.salary
LIMIT 1
)
要为每个部门得到的值,使用:
SELECT e.*
FROM department d
LEFT JOIN
employee e
ON e.id =
(
SELECT id
FROM employee ei
WHERE ei.department = d.id
ORDER BY
e.salary
LIMIT 1
)
要只对那些有雇员的部门获取值,使用:
SELECT e.*
FROM (
SELECT DISTINCT eo.department
FROM employee eo
) d
JOIN
employee e
ON e.id =
(
SELECT id
FROM employee ei
WHERE ei.department = d.department
ORDER BY
e.salary
LIMIT 1
)
当然,具有上(department, salary)
索引将大大提高所有三个查询。
最快溶液:
SET @dep := '';
SELECT * FROM (
SELECT * FROM `employee` ORDER BY `department`, `salary`
) AS t WHERE IF ( @dep = t.`department`, FALSE, ( @dep := t.`department` ) OR TRUE );
另一种方法可以是使用分析函数。下面是使用的分析和功能ROW_NUM查询
SELECT名字,薪水从(SELECT名字,薪水,ROW_NUMBER()以上(PARTITION BY DEPARTMENT_ID ORDER BY薪金ASC)从员工ROW_COUNT)其中ROW_COUNT = 1;
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