mysql - 每月和每年的总和
mysql - sums per months and years
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", "2020-03-01", "3", "700", "70", "1");
sql:
SELECT
YEAR(app_date) AS aYear,
Month(app_date) AS aMonth,
SUM(CASE WHEN work_id = 21 THEN app_price END) AS Work,
SUM(CASE WHEN work_id = 21 THEN app_price_paid END) AS Paid,
SUM(CASE WHEN work_id = 21 THEN app_price END)-SUM(CASE WHEN work_id = 21 THEN app_price_paid END) AS Debt
FROM tblappointment
GROUP BY YEAR(app_date), Month(app_date)
ORDER BY YEAR(app_date), Month(app_date);
现在输出:
YEAR MONTH WORK PAID DEBT
2020 1 numbers here...
2020 2 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....
谢谢
您可以使用 union all
取消透视,然后使用条件聚合进行透视:
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
) t
group by yr, details
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", "2020-03-01", "3", "700", "70", "1");
sql:
SELECT
YEAR(app_date) AS aYear,
Month(app_date) AS aMonth,
SUM(CASE WHEN work_id = 21 THEN app_price END) AS Work,
SUM(CASE WHEN work_id = 21 THEN app_price_paid END) AS Paid,
SUM(CASE WHEN work_id = 21 THEN app_price END)-SUM(CASE WHEN work_id = 21 THEN app_price_paid END) AS Debt
FROM tblappointment
GROUP BY YEAR(app_date), Month(app_date)
ORDER BY YEAR(app_date), Month(app_date);
现在输出:
YEAR MONTH WORK PAID DEBT
2020 1 numbers here...
2020 2 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....
谢谢
您可以使用 union all
取消透视,然后使用条件聚合进行透视:
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
) t
group by yr, details