Question

我正在尝试从列表中找到每个部门的第n大薪水。我可以使用聚合来执行基本的最小值和最大值，例如

Select DEPARTMENT, MAX(SALARY) FROM dept_salary
GROUP BY 1

使用MySQL我可以根据个人而不是合并来获得第n个工资：

SELECT DISTINCT DEPARTMENT, SALARY FROM dept_salary
ORDER BY 2 DESC
LIMIT n,1

DEPARTMENT         SALARY
-------------------------
Customer Service   143937
Human Resources    141953
Customer Service   138637
Customer Service   137535
Customer Service   136548
Customer Service   135650
Marketing          135497
Customer Service   134893
Customer Service   133837
Customer Service   133569

有关如何使其工作的任何指针都非常感谢。

Answer 1

对于一般解决方案，变量是最简单的方法：

this.iterationRecords = Ext.create('Rally.data.wsapi.Store', {
        model: 'Iteration',
        fetch: ["Name", START_DATE_FIELD, END_DATE_FIELD, "ObjectID", "State", "PlannedVelocity", "UserStories"], //Having trouble grabbing User Stories
            sorters: [
                {property: START_DATE_FIELD, direction: "DESC"},
                {property: END_DATE_FIELD, direction: "DESC"},
                {property: "State", operator: "=", value: "Accepted"}
            ],
        autoLoad: true,
        listeners: {
            load: function(store, records) {
                console.log(records);
            }
        }
    });

Answer 2

尝试按部门分组：

SELECT DEPARTMENT, MAX(SALARY) AS max_salary FROM dept_salary
GROUP BY DEPARTMENT
ORDER BY SALARY DESC

Answer 3

select 
 DISTINCT(salary) 
from dept_salary
 order by salary desc 
limit 1,1

Answer 4

set @nthMaxSalary = 2;

select ctr,dept,salary from (
select @rowCtr := if(@dept = dept,@rowCtr +1,1) as ctr
,@dept := dept dept
,salary
from dept_salary
order by dept,salary desc
  ) t where ctr = @nthMaxSalary ;

如何从不同部门找到第n个最高工资？

4 个答案: