我想通过ID将状态更改计数为
我有一个mysql状态表，该表的名称带有预实现，实现和操作状态。
所以我想计算这个月内所有已更新/更改其status_id的项目
-
CREATE TABLE `status` (
`status_id` int(11) NOT NULL,
`status_name` varchar(30) NOT NULL,
`description` varchar(255) DEFAULT NULL,
`status_name_tg` varchar(50) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `status` (`status_id`, `status_name`, `description`, `status_name_tg`) VALUES
(1, 'Pre Implementation', 'Operational', 'Pre Implementation'),
(2, 'Implementation', NULL, 'Implementation'),
(3, 'Operational', NULL, 'Operational'),
(4, 'Inactive', NULL, 'Inactive'),
(5, 'Cancellation', NULL, 'Cancellation');
CREATE TABLE `project_status` (
`project_status_id` int(11) NOT NULL,
`status_id` int(11) NOT NULL,
`time` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP,
`project_id` int(11) NOT NULL,
`reason_for_cancellation` varchar(255) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `project_status` (`project_status_id`, `status_id`, `time`, `project_id`, `reason_for_cancellation`) VALUES
(1, 1, '2019-02-24 21:51:50', 1, NULL),
(2, 2, '2019-03-26 21:52:57', 1, ' '),
(3, 1, '2019-04-30 21:57:57', 2, NULL),
(4, 1, '2019-05-26 22:04:08', 3, NULL),
(5, 3, '2019-08-24 22:06:36', 1, ' '),
(6, 2, '2019-08-11 22:07:05', 3, ' '),
(8, 1, '2019-08-01 00:14:41', 6, NULL),
(9, 1, '2019-08-09 12:11:22', 7, NULL),
(10, 1, '2019-08-09 12:15:22', 8, NULL),
(11, 3, '2019-08-14 10:07:49', 7, NULL),
(12, 2, '2019-08-14 10:10:45', 8, NULL),
(13, 2, '2019-08-26 17:16:02', 6, 'NULL');
(14, 3, '2019-08-26 17:16:02', 6, 'NULL');
CREATE TABLE `projects` (
`project_id` int(11) NOT NULL,
`name` varchar(150) NOT NULL,
`start_date` datetime NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `projects` (`project_id`, `name`, `start_date`) VALUES
(1, 'site A', '2019-02-01 00:00:00'),
(2, 'site B', '2019-03-12 00:00:00'),
(3, 'Site C', '2019-04-04 00:00:00'),
(4, 'Site D', '2019-05-03 00:00:00'),
(5, 'site E', '2019-06-01 00:00:00'),
(6, 'site F', '2019-08-02 00:00:00'),
(7, 'Site J', '2019-08-01 00:00:00'),
(8, 'Site H', '2019-08-05 00:00:00');
期望输出
|pre - implementations|implementations - operations| pre - operations|
|2 | 1 |2 |
答案 0 :(得分：1)
我认为关键是找出每个项目每个月的最新状态：
SELECT YEAR(time) as year
, MONTH(time) as month
, project_id
, MAX(status_id) as last_status
FROM project_status
WHERE status_id < 4 -- ignore (inactive, cancellation)
GROUP BY YEAR(time), MONTH(time), project_id;
输出
+------+-------+------------+-------------+
| year | month | project_id | last_status |
+------+-------+------------+-------------+
| 2019 | 2 | 1 | 1 |
| 2019 | 3 | 1 | 2 |
| 2019 | 4 | 2 | 1 |
| 2019 | 5 | 3 | 1 |
| 2019 | 8 | 1 | 3 |
| 2019 | 8 | 3 | 2 |
| 2019 | 8 | 6 | 2 |
| 2019 | 8 | 7 | 3 |
| 2019 | 8 | 8 | 2 |
+------+-------+------------+-------------+
现在，使用相关查询，您可以找到当月前的最后状态：
SELECT *, (SELECT MAX(p.status_id)
FROM project_status p
WHERE p.time < CONCAT(t.year,'/', t.month,'/1')
AND p.project_id = t.project_id
) as previous_status
FROM (
SELECT YEAR(time) as year
, MONTH(time) as month
, project_id
, MAX(status_id) as last_status
FROM project_status
WHERE status_id < 4
GROUP BY YEAR(time), MONTH(time), project_id
) t
输出
+------+-------+------------+-------------+-----------------+
| year | month | project_id | last_status | previous_status |
+------+-------+------------+-------------+-----------------+
| 2019 | 2 | 1 | 1 | |
| 2019 | 3 | 1 | 2 | 1 |
| 2019 | 4 | 2 | 1 | |
| 2019 | 5 | 3 | 1 | |
| 2019 | 8 | 1 | 3 | 2 |
| 2019 | 8 | 3 | 2 | 1 |
| 2019 | 8 | 6 | 2 | |
| 2019 | 8 | 7 | 3 | |
| 2019 | 8 | 8 | 2 | |
+------+-------+------------+-------------+-----------------+
现在只需进行一些条件计数
SELECT q.year
, q.month
, COUNT(CASE WHEN q.last_status = 2 THEN 1 END) as pre_implementation
, COUNT(CASE WHEN q.last_status = 3
AND q.previous_status = 2 THEN 1 END) as implementation_operation
, COUNT(CASE WHEN q.last_status = 3 THEN 1 END) as pre_operation
FROM (
SELECT *, (SELECT MAX(p.status_id)
FROM project_status p
WHERE p.time < CONCAT(t.year,'/', t.month,'/1')
AND p.project_id = t.project_id
) as previous_status
FROM (
SELECT YEAR(time) as year
, MONTH(time) as month
, project_id
, MAX(status_id) as last_status
FROM project_status
WHERE status_id < 4
GROUP BY YEAR(time), MONTH(time), project_id
) t
) q
GROUP BY q.year, q.month
输出
+------+-------+--------------------+--------------------------+---------------+
| year | month | pre_implementation | implementation_operation | pre_operation |
+------+-------+--------------------+--------------------------+---------------+
| 2019 | 2 | 0 | 0 | 0 |
| 2019 | 3 | 1 | 0 | 0 |
| 2019 | 4 | 0 | 0 | 0 |
| 2019 | 5 | 0 | 0 | 0 |
| 2019 | 8 | 3 | 1 | 2 |
+------+-------+--------------------+--------------------------+---------------+
答案 1 :(得分：0)
所以我想计算这个月内所有已更新/更改其status_id的项目
以下查询使您全部状态更改：
select status_name, prev_status_name, count(*)
from (select ps.*, s.status_name,
lag(s.status_name) over (partition by ps.project_id order by ps.time) as prev_status_name
from project_status ps join
status s
on ps.status_id = s.status_id
) ps
where ps.time >= curdate() - interval (1 - day(curdate())) day and
status_name <> prev_status_name
group by status_name, prev_status_name;
如果需要特定的过滤器，可以进一步过滤。
注意：此处的结果在单独的行中，而不是在单独的列中。
您可以使用以下方法在单独的列中获取结果：
select sum( prev_status_name = 'Pre Implementation' and status_name = 'Implementation' ) as cnt1,
. . . -- for the remaining pairs that you want
from (select ps.*, s.status_name,
lag(s.status_name) over (partition by ps.project_id order by ps.time) as prev_status_name
from project_status ps join
status s
on ps.status_id = s.status_id
) ps
where ps.time >= curdate() - interval (1 - day(curdate())) day and
status_name <> prev_status_name;
以上内容适用于最新版本的MySQL。在早期版本中，您可以使用相关的子查询或变量。例如：
select status_name, prev_status_name, count(*)
from (select ps.*, s.status_name,
(select s2.status_name
from project_status ps2 join
status s2
on ps2.status_id = s2.status_id
where ps2.project_id = ps.project_id and
ps2.time < ps.time
order by ps2.time desc
) as prev_status_name
from project_status ps join
status s
on ps.status_id = s.status_id
where ps.time >= curdate() - interval (1 - day(curdate())) day and
) ps
where status_name <> prev_status_name
group by status_name, prev_status_name;