mysql - 每年每行每 months/years 的总和
mysql - sums per months/years for every row per year
我有以下演示
CREATE TABLE `tblappointment` (
`app_id` mediumint(8) UNSIGNED NOT NULL,
`app_date` date NOT NULL,
`work_id` smallint(5) UNSIGNED NOT NULL,
`app_price` double DEFAULT NULL,
`app_price_paid` double DEFAULT NULL,
`receipt_id` tinyint(3) UNSIGNED DEFAULT NULL
);
INSERT INTO `tblappointment` (`app_id`, `app_date`, `work_id`, `app_price`, `app_price_paid`, `receipt_id`) VALUES
("1", "2020-03-01", "21", "100", "50", "1"),
("2", "2020-04-01", "21", "200", "40", "3"),
("4", "2020-06-01", "2", "500", "70", "1"),
("5", "2020-07-01", "21", "300", "30", "1"),
("6", "2020-08-01", "21", "200", "20", "2"),
("7", "2020-09-01", "5", "100", "50", "1"),
("8", "2020-10-01", "6", "200", "30", "2"),
("9", "2020-11-01", "21", "300", "30", "1"),
("10", "2020-12-01", "21", "400", "20", "3"),
("11", "2020-01-01", "8", "500", "90", "1"),
("12", "2020-02-01", "21", "600", "80", "5"),
("13", "2021-03-01", "3", "700", "70", "1");
sql:
select
year(app_date) yr,
details,
sum(case when month(app_date) = 1 then val else 0 end) month_01,
sum(case when month(app_date) = 2 then val else 0 end) month_02,
sum(case when month(app_date) = 3 then val else 0 end) month_03,
sum(case when month(app_date) = 4 then val else 0 end) month_04,
sum(case when month(app_date) = 5 then val else 0 end) month_05,
sum(case when month(app_date) = 6 then val else 0 end) month_06,
sum(case when month(app_date) = 7 then val else 0 end) month_07,
sum(case when month(app_date) = 8 then val else 0 end) month_08,
sum(case when month(app_date) = 9 then val else 0 end) month_09,
sum(case when month(app_date) = 10 then val else 0 end) month_10,
sum(case when month(app_date) = 11 then val else 0 end) month_11,
sum(case when month(app_date) = 22 then val else 0 end) month_12
from (
select app_date, app_price val, 'work' details from tblappointment where work_id = 21
union all
select app_date, app_price_paid val, 'paid' details from tblappointment where work_id = 21
union all
select app_date, app_price - app_price_paid val, 'debt' details from tblappointment where work_id = 21
union all
select app_date, app_price val, 'test' details from tblappointment where work_id = 3
) t
group by yr, details
现在输出：
YEAR DETAILS 1 2 3 4 5 6 7 8 9 10 11 12
2020 Work numbers here....
2020 Paid numbers here....
2020 Debt numbers here....
2021 Test numbers here....
我想要这样的输出：
YEAR DETAILS 1 2 3 4 5 6 7 8 9 10 11 12
2020 Work numbers here....
2020 Paid numbers here....
2020 Debt numbers here....
2020 Test NULL or O or Empty
2020 Work NULL or O or Empty
2020 Paid NULL or O or Empty
2020 Debt NULL or O or Empty
2021 Test numbers here....
如您所见，'Test' 没有 2020 年的值，只有 2021 年的值。
其余的具有 2020 年的值，但没有 2021 年的数据。
每年都需要显示所有 'details'，要么是 NULL，要么是 0，要么是 Empty。
谢谢。
您可以加入不同的年份和所有详细信息，然后加入您的查询：
select
y.yr,
d.details,
sum(case when month(app_date) = 1 then val else 0 end) month_01,
sum(case when month(app_date) = 2 then val else 0 end) month_02,
sum(case when month(app_date) = 3 then val else 0 end) month_03,
sum(case when month(app_date) = 4 then val else 0 end) month_04,
sum(case when month(app_date) = 5 then val else 0 end) month_05,
sum(case when month(app_date) = 6 then val else 0 end) month_06,
sum(case when month(app_date) = 7 then val else 0 end) month_07,
sum(case when month(app_date) = 8 then val else 0 end) month_08,
sum(case when month(app_date) = 9 then val else 0 end) month_09,
sum(case when month(app_date) = 10 then val else 0 end) month_10,
sum(case when month(app_date) = 11 then val else 0 end) month_11,
sum(case when month(app_date) = 22 then val else 0 end) month_12
from (
select 'work' details union all select 'paid' union all
select 'debt' details union all select 'test' details
) d cross join (
select distinct year(app_date) yr
from tblappointment
where work_id in (3, 21)
) y
left join (
select app_date, app_price val, 'work' details from tblappointment where work_id = 21
union all
select app_date, app_price_paid val, 'paid' details from tblappointment where work_id = 21
union all
select app_date, app_price - app_price_paid val, 'debt' details from tblappointment where work_id = 21
union all
select app_date, app_price val, 'test' details from tblappointment where work_id = 3
) t on year(t.app_date) = y.yr and t.details = d.details
group by y.yr, d.details
参见demo。
结果：
> yr | details | month_01 | month_02 | month_03 | month_04 | month_05 | month_06 | month_07 | month_08 | month_09 | month_10 | month_11 | month_12
> ---: | :------ | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------:
> 2020 | debt | 0 | 520 | 50 | 160 | 0 | 0 | 270 | 180 | 0 | 0 | 270 | 0
> 2020 | paid | 0 | 80 | 50 | 40 | 0 | 0 | 30 | 20 | 0 | 0 | 30 | 0
> 2020 | test | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
> 2020 | work | 0 | 600 | 100 | 200 | 0 | 0 | 300 | 200 | 0 | 0 | 300 | 0
> 2021 | debt | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
> 2021 | paid | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
> 2021 | test | 0 | 0 | 700 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
> 2021 | work | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
我有以下演示
CREATE TABLE `tblappointment` (
`app_id` mediumint(8) UNSIGNED NOT NULL,
`app_date` date NOT NULL,
`work_id` smallint(5) UNSIGNED NOT NULL,
`app_price` double DEFAULT NULL,
`app_price_paid` double DEFAULT NULL,
`receipt_id` tinyint(3) UNSIGNED DEFAULT NULL
);
INSERT INTO `tblappointment` (`app_id`, `app_date`, `work_id`, `app_price`, `app_price_paid`, `receipt_id`) VALUES
("1", "2020-03-01", "21", "100", "50", "1"),
("2", "2020-04-01", "21", "200", "40", "3"),
("4", "2020-06-01", "2", "500", "70", "1"),
("5", "2020-07-01", "21", "300", "30", "1"),
("6", "2020-08-01", "21", "200", "20", "2"),
("7", "2020-09-01", "5", "100", "50", "1"),
("8", "2020-10-01", "6", "200", "30", "2"),
("9", "2020-11-01", "21", "300", "30", "1"),
("10", "2020-12-01", "21", "400", "20", "3"),
("11", "2020-01-01", "8", "500", "90", "1"),
("12", "2020-02-01", "21", "600", "80", "5"),
("13", "2021-03-01", "3", "700", "70", "1");
sql:
select
year(app_date) yr,
details,
sum(case when month(app_date) = 1 then val else 0 end) month_01,
sum(case when month(app_date) = 2 then val else 0 end) month_02,
sum(case when month(app_date) = 3 then val else 0 end) month_03,
sum(case when month(app_date) = 4 then val else 0 end) month_04,
sum(case when month(app_date) = 5 then val else 0 end) month_05,
sum(case when month(app_date) = 6 then val else 0 end) month_06,
sum(case when month(app_date) = 7 then val else 0 end) month_07,
sum(case when month(app_date) = 8 then val else 0 end) month_08,
sum(case when month(app_date) = 9 then val else 0 end) month_09,
sum(case when month(app_date) = 10 then val else 0 end) month_10,
sum(case when month(app_date) = 11 then val else 0 end) month_11,
sum(case when month(app_date) = 22 then val else 0 end) month_12
from (
select app_date, app_price val, 'work' details from tblappointment where work_id = 21
union all
select app_date, app_price_paid val, 'paid' details from tblappointment where work_id = 21
union all
select app_date, app_price - app_price_paid val, 'debt' details from tblappointment where work_id = 21
union all
select app_date, app_price val, 'test' details from tblappointment where work_id = 3
) t
group by yr, details
现在输出：
YEAR DETAILS 1 2 3 4 5 6 7 8 9 10 11 12
2020 Work numbers here....
2020 Paid numbers here....
2020 Debt numbers here....
2021 Test numbers here....
我想要这样的输出：
YEAR DETAILS 1 2 3 4 5 6 7 8 9 10 11 12
2020 Work numbers here....
2020 Paid numbers here....
2020 Debt numbers here....
2020 Test NULL or O or Empty
2020 Work NULL or O or Empty
2020 Paid NULL or O or Empty
2020 Debt NULL or O or Empty
2021 Test numbers here....
如您所见，'Test' 没有 2020 年的值，只有 2021 年的值。 其余的具有 2020 年的值，但没有 2021 年的数据。 每年都需要显示所有 'details'，要么是 NULL，要么是 0，要么是 Empty。
谢谢。
您可以加入不同的年份和所有详细信息，然后加入您的查询：
select
y.yr,
d.details,
sum(case when month(app_date) = 1 then val else 0 end) month_01,
sum(case when month(app_date) = 2 then val else 0 end) month_02,
sum(case when month(app_date) = 3 then val else 0 end) month_03,
sum(case when month(app_date) = 4 then val else 0 end) month_04,
sum(case when month(app_date) = 5 then val else 0 end) month_05,
sum(case when month(app_date) = 6 then val else 0 end) month_06,
sum(case when month(app_date) = 7 then val else 0 end) month_07,
sum(case when month(app_date) = 8 then val else 0 end) month_08,
sum(case when month(app_date) = 9 then val else 0 end) month_09,
sum(case when month(app_date) = 10 then val else 0 end) month_10,
sum(case when month(app_date) = 11 then val else 0 end) month_11,
sum(case when month(app_date) = 22 then val else 0 end) month_12
from (
select 'work' details union all select 'paid' union all
select 'debt' details union all select 'test' details
) d cross join (
select distinct year(app_date) yr
from tblappointment
where work_id in (3, 21)
) y
left join (
select app_date, app_price val, 'work' details from tblappointment where work_id = 21
union all
select app_date, app_price_paid val, 'paid' details from tblappointment where work_id = 21
union all
select app_date, app_price - app_price_paid val, 'debt' details from tblappointment where work_id = 21
union all
select app_date, app_price val, 'test' details from tblappointment where work_id = 3
) t on year(t.app_date) = y.yr and t.details = d.details
group by y.yr, d.details
参见demo。
结果：
> yr | details | month_01 | month_02 | month_03 | month_04 | month_05 | month_06 | month_07 | month_08 | month_09 | month_10 | month_11 | month_12
> ---: | :------ | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------: | -------:
> 2020 | debt | 0 | 520 | 50 | 160 | 0 | 0 | 270 | 180 | 0 | 0 | 270 | 0
> 2020 | paid | 0 | 80 | 50 | 40 | 0 | 0 | 30 | 20 | 0 | 0 | 30 | 0
> 2020 | test | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
> 2020 | work | 0 | 600 | 100 | 200 | 0 | 0 | 300 | 200 | 0 | 0 | 300 | 0
> 2021 | debt | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
> 2021 | paid | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
> 2021 | test | 0 | 0 | 700 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
> 2021 | work | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0