Question

我需要总结与有效日期的参考文献相关的数据。

让我们想象一下这些表的工资架构

employee                                         work
+----+------+                                    +----+--------+------------+
| id | name |                                    | id | emp_id | date       |
+----+------+                                    +----+--------+------------+
|  1 | John |                                    |  1 |      1 | 2012-10-05 |
|  2 | Bob  |                                    |  2 |      1 | 2012-02-10 |
+----+------+                                    |  3 |      2 | 2012-09-03 |
                                                 |  4 |      1 | 2012-12-12 |
salary                                           |  5 |      1 | 2012-04-04 |
+----+--------+--------------+------------+      |  6 |      2 | 2012-06-09 |
| id | emp_id | daily_amount | start_date |      |  7 |      1 | 2012-07-24 |
+----+--------+--------------+------------+      +----+--------+------------+
|  1 |      1 |           10 | 2012-01-01 | 
|  2 |      1 |           15 | 2012-04-16 |
|  3 |      2 |           20 | 2012-05-25 |
+----+--------+--------------+------------+

从2012-01-01 约翰每天支付10天，然后从2012-04-16，他的工资增加到15。 Bob 自2012-05-25以来已支付20美元

现在让我们说我需要计算2012年两名员工的工资。这是我天真的尝试：

SELECT
  emp.name AS "employee"
, SUM(sal.daily_amount) AS "amount"
FROM employee AS emp
JOIN work AS wrk ON (wrk.emp_id = emp.id)
JOIN salary AS sal ON (
  sal.emp_id = emp.id
  AND wrk.date >= sal.start_date
)
WHERE YEAR(wrk.date) = 2012
GROUP BY emp.id
;

当然它没有给我我想要的东西，因为它没有取得MAX start_date

问题：Bob＆amp; S获得的金额是多少？约翰在2012年？

Answer 1

在SQLfiddle上测试了以下内容：

SELECT employee.name AS "employee",
       SUM(s1.daily_amount) AS "amount"
FROM work
  INNER JOIN salary s1
     ON work.emp_id = s1.emp_id
    AND work.date >= s1.start_date
  LEFT JOIN salary s2
     ON work.emp_id = s2.emp_id
    AND work.date >= s2.start_date
    AND s2.start_date > s1.start_date
  INNER JOIN employee
     ON work.emp_id = employee.id
WHERE s2.emp_id IS NULL
GROUP BY work.emp_id

我们的想法是使用s1和s2的组合来查找上一个日期：s1是工作日之前的任何薪资定义，而{{1}只有在没有适用的后期工资定义的情况下才是s2。

Answer 2

嗯，你有salary table的开始日期，但你也需要一个结束日期。您可以使用派生表执行此操作：

  SELECT
    s_1.emp_id,
    s_1.daily_amount,
    s_1.start_date AS start,
    DATE_ADD(MIN(s_2.start_date), INTERVAL -1 DAY) AS end
  FROM
    salary s_1
    LEFT JOIN salary s_2
      ON s_1.emp_id = s_2.emp_id AND s_1.start_date < s_2.start_date
  GROUP BY s_1.emp_id, s_1.start_date

最后看起来像：

SELECT
  employee.name,
  SUM(s.daily_amount) AS amount
FROM
  employee
  JOIN work
    ON employee.id = work.emp_id
  JOIN
    (
      SELECT
        s_1.emp_id,
        s_1.daily_amount,
        s_1.start_date AS start,
        DATE_ADD(MIN(s_2.start_date), INTERVAL -1 DAY) AS end
      FROM
        salary s_1
        LEFT JOIN salary s_2
          ON s_1.emp_id = s_2.emp_id AND s_1.start_date < s_2.start_date
      GROUP BY s_1.emp_id, s_1.start_date
    ) AS s
    ON employee.id = s.emp_id
    AND
    work.date >= s.start
    AND
    (work.date <= s.end OR s.end IS NULL)
GROUP BY employee.id

修改没有DATE_ADD()

SELECT employee.name, SUM(s.daily_amount) AS amount FROM employee JOIN work ON employee.id = work.emp_id JOIN ( SELECT s_1.emp_id, s_1.daily_amount, s_1.start_date AS start, MIN(s_2.start_date) AS end FROM salary s_1 LEFT JOIN salary s_2 ON s_1.emp_id = s_2.emp_id AND s_1.start_date < s_2.start_date GROUP BY s_1.emp_id, s_1.start_date ) AS s ON employee.id = s.emp_id AND work.date >= s.start AND (work.date < s.end OR s.end IS NULL) GROUP BY employee.id

MySQL - 加入有效的日期数据

2 个答案: