Question

我有一个记录项目状态的表：

CREATE TABLE `project_states` (
    `id` INT(10) NOT NULL AUTO_INCREMENT,
    `state_id` INT(10) NULL DEFAULT NULL,
    `project_id` INT(10) NULL DEFAULT NULL,
    `created_at` TIMESTAMP NULL DEFAULT NULL,
    PRIMARY KEY (`id`)
)
COLLATE='utf8_general_ci'
ENGINE=InnoDB

一些假数据：

    id  | state_id | project_id | created_at
-----------------------------------------------------
     1  |   2      |   8        | 2014-05-27 10:58:12   
     2  |   3      |   8        | 2014-05-27 11:10:34
     3  |   8      |   8        | 2014-05-27 11:56:48
     4  |   2      |  10        | 2014-05-27 11:08:34
     5  |   4      |  10        | 2014-05-27 11:59:01

我试图获得两个州之间的时差（比如说2和8），并且只针对确实这两个州的项目;在这种情况下，仅适用于项目8（因为10没有状态8）。

到目前为止，我完全设法选择符合条件的项目（包含两个值，而不仅仅是一个），但此查询返回每个匹配项目的结果元组：

SELECT t.*
   FROM (
     SELECT ps.*
     FROM project_states ps
     WHERE ps.state_id IN (2,8) 
   ) as t
JOIN project_states pro ON pro.project_id = t.project_id
WHERE pro.state_id = 8

正确返回：

    id  | state_id | project_id | created_at
-----------------------------------------------------
     1  |   2      |   8        | 2014-05-27 10:58:12   
     3  |   8      |   8        | 2014-05-27 11:56:48

我确信它有效，因为如果我将缺少的状态添加到另一个项目中，它会返回新的结果元组：

    id  | state_id | project_id | created_at
-----------------------------------------------------
     1  |   2      |   8        | 2014-05-27 10:58:12   
     3  |   8      |   8        | 2014-05-27 11:56:48
     4  |   2      |  10        | 2014-05-27 11:08:34
     6  |   8      |  10        | 2014-05-27 12:03:08

但我如何计算时差呢？我使用PHP，我知道我可以通过project_id循环结果，然后计算差异，但我认为可能有一个纯SQL解决方案会产生如下结果：< / p>

project_id | difference
------------------------
     8     | 0000-00-00 01:02:00
    10     | 0000-00-00 01:05:26

嗯，实际上我的目标是计算项目位于这两个选定状态之间的平均时差，因此所有记录都可以归结为一个平均值，但可能我后来才知道这个问题。

Answer 1

我会使用aggregation和having子句来执行此操作：

select ps.project_id,
       timestampdiff(second, min(case when ps.state_id = 2 then created_at end),
                     max(case when ps.state_id = 8 then created_at)
                    ) as diff_in_secs

from project_states ps
where ps.state_id in (2, 8)
group by ps.project_id
having count(distinct ps.state_id) = 2;

您也可以使用连接执行此操作：

select ps2.project_id, timestampdiff(second, ps2.created_at, ps8.created_at)
from project_states ps2 join
     project_states ps8
     on ps2.project_id = ps8.project_id and
        ps2.state_id = 2 and ps8.state_id = 8;

Answer 2

如果您只有2个状态，则可以使用简单的INNER JOIN。并且TIMESTAMPDIFF()可以在几分钟内获得差异，例如：

SELECT p.project_id,
       TIMESTAMPDIFF(MINUTE,p.created_at,p1.created_at) 
          as state_minutes_diff

FROM project_states as p
JOIN project_states as p1 
  ON p.project_id=p1.project_id
     AND p1.state_id=8  

WHERE p.state_id=3

SQLFiddle demo

获得所有项目的平均值：

SELECT AVG(TIMESTAMPDIFF(MINUTE,p.created_at,p1.created_at))
          as AVG_state_minutes_diff
FROM project_states as p
JOIN project_states as p1 
  ON p.project_id=p1.project_id
     AND p1.state_id=8  
WHERE p.state_id=3

SQLFiddle demo

Answer 3

实际上，您不需要JOIN，因为您可以通过指定正确的GROUP BY子句来解决问题：

SELECT
  SEC_TO_TIME(AVG(times.atime)) AS avg_time
FROM
  (SELECT
    project_id,
    TIMEDIFF(MAX(created_at), MIN(created_at)) AS atime
  FROM
    project_states
  WHERE
    state_id IN (2,8)
  GROUP BY
    project_id
  HAVING
    COUNT(DISTINCT id)>1) as times

_{我似乎迟到了，但至少，因为fiddle我做了}

Answer 4

SELECT project_id,
       timediff(min(created_at), max(created_at)) as difference
FROM project_states
WHERE state_id IN (2,8) 
GROUP BY project_id 
having count(distinct state_id) = 2

TIMEDIFF（）在具有一个值和另一个值的记录之间，而不仅仅是一个

4 个答案: