我有一个 employees 表，其中每个雇员都有一个相关的开始日期，结束日期和一个薪金。 / p>
注意：：在底部，您可以找到用于导入结构和数据的SQL代码。
+----+-------+------------+------------+---------+
| id | name | start_date | end_date | salary |
+----+-------+------------+------------+---------+
| 1 | Mark | 2017-05-01 | 2020-01-31 | 2000.00 |
| 2 | Tania | 2018-02-01 | 2019-08-31 | 5000.00 |
| 3 | Leo | 2018-02-01 | 2018-09-30 | 3000.00 |
| 4 | Elsa | 2018-12-01 | 2020-05-31 | 4000.00 |
+----+-------+------------+------------+---------+
对于给定的日期范围，我想提取给定日期范围内每个月的平均工资。
更新：我想拥有MySQL 5.6的解决方案，但是也有适用于MySQL 8+的解决方案（仅出于个人知识）将非常棒。
如果日期范围为 2018-08-01 - 2019-01-31 ，则SQL报表应在2018年8月至2019年1月之间循环，并且必须计算每个月的平均工资：
下面您将看到日期范围 2018-08-01 - 2019-01-31 < / strong>
+------+-------+------------+
| year | month | avg_salary |
+------+-------+------------+
| 2018 | 08 | 3333.33 |
| 2018 | 09 | 3333.33 |
| 2018 | 10 | 3500.00 |
| 2018 | 11 | 3500.00 |
| 2018 | 12 | 3666.67 |
| 2019 | 01 | 3666.67 |
+------+-------+------------+
注意：我解决了将MySQL与PHP代码混合使用的问题，但在较大的日期范围内，它必须执行过多的查询（每个月执行一次）。因此，我想有一个仅使用MySQL 的解决方案。
CREATE TABLE `employees` (
`id` int(10) UNSIGNED NOT NULL PRIMARY KEY AUTO_INCREMENT,
`name` varchar(50) NOT NULL,
`start_date` date NOT NULL,
`end_date` date NOT NULL,
`salary` decimal(10,2) DEFAULT NULL
);
INSERT INTO `employees` (`id`, `name`, `start_date`, `end_date`, `salary`) VALUES
(1, 'Mark', '2017-05-01', '2020-01-31', '2000.00'),
(2, 'Tania', '2018-02-01', '2019-08-31', '5000.00'),
(3, 'Leo', '2018-02-01', '2018-09-30', '3000.00'),
(4, 'Elsa', '2018-12-01', '2020-05-31', '4000.00');
答案 0 :(得分：3)
这是MySQL 8.0递归CTE的实现方式。 CTE在year, month表中的最小start_date和最大end_date之间创建所有employees组合的列表，然后将其LEFT JOIN固定到{ {1}}表以获取该特定年份和月份中所有在职员工的平均工资：
employees输出：
WITH RECURSIVE months (year, month) AS
(
SELECT YEAR(MIN(start_date)) AS year, MONTH(MIN(start_date)) AS month FROM employees
UNION ALL
SELECT year + (month = 12), (month % 12) + 1 FROM months
WHERE STR_TO_DATE(CONCAT_WS('-', year, month, '01'), '%Y-%m-%d') <= (SELECT MAX(end_date) FROM employees)
)
SELECT m.year, m.month, ROUND(AVG(e.salary), 2) AS avg_salary
FROM months m
LEFT JOIN employees e ON STR_TO_DATE(CONCAT_WS('-', m.year, m.month, '01'), '%Y-%m-%d') BETWEEN e.start_date AND e.end_date
WHERE STR_TO_DATE(CONCAT_WS('-', m.year, m.month, '01'), '%Y-%m-%d') BETWEEN '2018-08-01' AND '2019-01-31'
GROUP BY m.year, m.month
答案 1 :(得分：2)
您只需键入所需的月份（或使用PHP代码生成它们）并加入：
SELECT ym, AVG(salary)
FROM (
SELECT '2018-08-01' + INTERVAL 0 MONTH AS ym UNION ALL
SELECT '2018-08-01' + INTERVAL 1 MONTH UNION ALL
SELECT '2018-08-01' + INTERVAL 2 MONTH UNION ALL
SELECT '2018-08-01' + INTERVAL 3 MONTH UNION ALL
SELECT '2018-08-01' + INTERVAL 4 MONTH UNION ALL
SELECT '2018-08-01' + INTERVAL 5 MONTH
) AS yearmonths
JOIN employees ON ym BETWEEN start_date AND end_date
GROUP BY ym
如果您的表包含数字0、1 ...，则可以使用该表。您甚至可以使用具有足够行数的 任何 表：
SELECT ym, AVG(salary)
FROM (
SELECT '2018-08-01' + INTERVAL @n := @n + 1 MONTH AS ym
FROM anytable, (SELECT @n := -1) x
LIMIT 100
) AS yearmonths
JOIN employees ON ym BETWEEN start_date AND end_date
WHERE ym <= '2019-01-01'
GROUP BY ym
答案 2 :(得分：1)
要完成此操作，您需要从日期范围生成天列表。这是一个关于SO的常见问题，我使用了this post中公认的解决方案。它使用一种简单的算术方法，并且可以生成日期的广泛列表（尽管性能可能会受到影响）。
然后，我们只需要与原始表进行联接，以计算该时间点的平均工资。
select
year(x.date),
month(x.date),
avg(coalesce(e.salary, 0)) avg_salary
from (
select a.date
from (
select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a) + (1000 * d.a) ) DAY as Date
from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as d
) a
where a.date between '2018-08-01' and '2019-01-31'
) x left join employees e ON x.date between e.start_date and e.end_date
group by year(x.date), month(x.date)
order by 1, 2
| year(x.date) | month(x.date) | avg_salary |
| ------------ | ------------- | ----------- |
| 2018 | 8 | 3333.333333 |
| 2018 | 9 | 3333.333333 |
| 2018 | 10 | 3500 |
| 2018 | 11 | 3500 |
| 2018 | 12 | 3666.666667 |
| 2019 | 1 | 3666.666667 |
PS：另一种方法是创建一个日历表，该表存储天数列表，然后只是：
select
year(x.date),
month(x.date),
avg(coalesce(e.salary, 0)) avg_salary
from
mycalendar x
left join employees e ON x.date between e.start_date and e.end_date
where x.date between '2018-08-01' and '2019-01-31'
group by year(x.date), month(x.date)
order by 1, 2
答案 3 :(得分：0)
部分答案...
这是一个使用整数（0-9）表的“老派”解决方案，但请注意，这种情况在较新版本的sql中是多余的...
SELECT * FROM ints;
+---+
| i |
+---+
| 0 |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 8 |
| 9 |
+---+
SELECT '2018-08-01' + INTERVAL i2.i * 10 + i1.i MONTH x
FROM ints i1
, ints i2
WHERE '2018-08-01' + INTERVAL i2.i * 10 + i1.i MONTH BETWEEN '2018-08-01' AND '2019-01-31';
+------------+
| x |
+------------+
| 2018-08-01 |
| 2018-09-01 |
| 2018-10-01 |
| 2018-11-01 |
| 2018-12-01 |
| 2019-01-01 |
+------------+
答案 4 :(得分：-1)
以下是执行此操作的Postgresql方法。可以通过更改Mysql中的generate_series() link和Extract()的等效项将其转换为Mysql查询
WITH cte1 AS
(SELECT generate_series('2018-08-01', '2019-01-31', '1 month'::interval)::date AS date),
cte2 AS
(SELECT id,
name,
salary,
generate_series(start_date, end_date, '1 month'::interval)::date AS date
FROM employees)
SELECT extract(YEAR
FROM cte1.date),
extract(MONTH
FROM cte1.date),
avg(salary)
FROM cte1
JOIN cte2 ON extract(MONTH
FROM cte1.date)=extract(MONTH
FROM cte2.date)
AND extract(YEAR
FROM cte1.date)=extract(YEAR
FROM cte2.date)
GROUP BY extract(YEAR
FROM cte1.date),
extract(MONTH
FROM cte1.date);