SQL

Question

这是表的 UML：https://dev.mysql.com/doc/employee/en/sakila-structure.html

这是我的尝试：

SELECT
    CONCAT(employees.first_name, ' ', employees.last_name) AS 'EmployeeName',
    salaries.emp_no AS 'Employee Number',
    departments.dept_no AS 'Department Number',
    departments.dept_name AS 'Department name',
    AVG(salaries.salary) AS 'Average salary'
FROM salaries
INNER JOIN dept_emp
    ON salaries.emp_no = dept_emp.emp_no
INNER JOIN employees
    ON salaries.emp_no = employees.emp_no
INNER JOIN departments
    ON dept_emp.dept_no = departments.dept_no
GROUP BY    salaries.emp_no,
            dept_emp.dept_no

我只是得到一个人工作过的所有部门的平均员工工资。我想要的结果必须按每个部门的每个员工分组。感谢任何帮助。

Answer 1

你可以试试这个查询

SELECT
    CONCAT(employees.first_name, ' ', employees.last_name) AS 'EmployeeName',
    salaries.emp_no AS 'Employee Number',
    departments.dept_no AS 'Department Number',
    departments.dept_name AS 'Department name',
    Cte_DepartmentSalaries.AvgSalary AS 'Average Salary'

--AVG(salaries.salary) AS 'Average salary'

FROM salaries
INNER JOIN dept_emp
    ON salaries.emp_no = dept_emp.emp_no
INNER JOIN employees
    ON salaries.emp_no = employees.emp_no
INNER JOIN departments
    ON dept_emp.dept_no = departments.dept_no   
LEFT JOIN (SELECT
    departments.dept_no,
    departments.dept_name,
    AVG(Salaries.Salary) AS AvgSalary
FROM Salaries
INNER JOIN dept_emp
    ON salaries.emp_no = dept_emp.emp_no
INNER JOIN departments
    ON dept_emp.dept_no = departments.dept_no
GROUP BY    departments.dept_no,
            departments.dept_name) AS Cte_DepartmentSalaries
    ON dept_emp.dept_no = Cte_DepartmentSalaries.dept_no

Answer 2

将您的群组更改为

GROUP BY employees.first_name, employees.last_name,  
         salaries.emp_no, departments.dept_no, departments.dept_name

mysql 允许您（恕我直言，以一种无用的方式）不将所有列用于分组依据。与其他平台不同，如果它们不在聚合函数中，则强制您将它们全部放入。所以你得到了不止一个同一行，但你无法分辨，因为 mysql 为你压扁了它们。

您可能在加入时也需要这个

INNER JOIN dept_emp ON salaries.emp_no = dept_emp.emp_no
  AND salaries.from_date = dept_emp.from_date 
  AND salaries.to_date = dept_emp.to_date

Answer 3

如果您加入 to_date 从日期到现在之间的薪水，您将获得这些时间段内薪水的任何变化。

SELECT  CONCAT(e.first_name,' ',e.last_name) AS 'EmployeeName',
        s.emp_no AS 'Employee Number',
        d.dept_no AS 'Department Number',
        d.dept_name AS 'Department name',
        AVG(s.salaries.salary) AS 'Average salary'
FROM    employees e
        JOIN dept_emp de ON e.emp_no = de.emp_no
        JOIN salaries s ON s.emp_no = de.emp_no
                           AND (de.to_date >= s.from_date AND de.to_date <= s.to_date)
        JOIN departments d ON d.dept_no = de.dept_no
GROUP BY s.emp_no,
        d.dept_no

Answer 4

看起来您需要检查员工工资历史记录 (salaries table) 和员工部门的 from_date 和 to_date 是否存在重叠历史 (dept_emp table).

假设 NULL 的 to_date 值表示 "current"，要找到重叠，连接将是这样的：

 FROM salaries s 
 JOIN dept_emp de
   ON ( de.emp_no    = s.emp_no )
  AND ( de.from_date <= s.to_date   OR s.to_date  IS NULL )  
  AND ( de.to_date   >= s.from_date OR de.to_date IS NULL )

就获得平均工资而言，这会稍微复杂一些，具体取决于您如何定义 "average"。

举个例子，某部门员工，一年年工资2万，然后四年[=35年工资3万=] 年...平均工资应该报告为 25,000（两个不同值的平均值），还是 28,000（所有年份的总工资除以年数。）

为了得到后一个结果，我们可以使用一个表达式来计算工资在部门中有效的天数...

TIMESTAMPDIFF(DAY ,GREATEST(de.from_date,s.from_date) ,LEAST(IFNULL(de.to_date,DATE(NOW())),IFNULL(s.to_date,DATE(NOW()))) )

我们可以将其乘以有效工资，将所有这些加在一起，然后除以总天数：

SUM(days*salary)/SUM(days)

类似这样的表达式：

SUM( TIMESTAMPDIFF(DAY ,GREATEST(de.from_date,s.from_date) ,LEAST(IFNULL(de.to_date,DATE(NOW())),IFNULL(s.to_date,DATE(NOW()))) ) * s.salary ) / SUM( TIMESTAMPDIFF(DAY ,GREATEST(de.from_date,s.from_date) ,LEAST(IFNULL(de.to_date,DATE(NOW())),IFNULL(s.to_date,DATE(NOW()))) ) )

Answer 5

要仅查找 table 的平均工资，请使用此 mySQL 查询。

SELECT AVG(salary), COUNT(*) 
     FROM employees;MY

Answer 6

SELECT AVG（工资）来自员工分组依据（departments_id）；

SQL - 如何获取每个部门的员工平均工资？

SQL - How can I get the average employee salary for each department?

mysql

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