Question

当我对数据进行分组时，我想计算日期 - 时间差异。

让我们看一下表格内容示例：

id | session_id | created
1  | 101        | 1/10/2010 9:30:10
2  | 102        | 1/10/2010 9:31:10
3  | 101        | 1/10/2010 9:32:10
4  | 103        | 1/10/2010 9:35:10
5  | 102        | 1/10/2010 9:38:10
6  | 103        | 1/10/2010 9:39:10

输出应如下：

session_id | time_count
101        | 2 (minutes)
102        | 7 (minutes)
103        | 4 (minutes)

所以我想在这里计算出来的＆＃39;时间差。数据字段，按session_id分组。

任何帮助都会很棒。在此先感谢：）

我的情况：

我有一个包含以下字段的表格：

id, session_id, platform, event, created

有2个事件（Start,End）

现在，如果有相同session_id的开始和结束记录，那么我想要完成此会话所需的时间。

但是，如果我有2个开始会话，但只有1个结束会话，那么我不希望第2个会话的时间差，因为它没有结束

Answer 1

SELECT session_id,
DATEDIFF(MINUTE, MAX(created), MIN(created)) AS diff
FROM table
GROUP BY session_id

Answer 2

试试这个：

表架构：

CREATE TABLE A(id INT, session_id INT,Event VARCHAR(20), created DATETIME);

INSERT INTO A VALUES(1, 101,'Start','2010/10/01 9:30:10')
,(2, 102,'Start'  , '2010/10/01 9:31:10')
,(3, 101,'End'    , '2010/10/01 9:32:10')
,(4, 103,'Start'  , '2010/10/01 9:35:10')
,(5, 102,'End'    , '2010/10/01 9:38:10')
,(6, 103,'End'    , '2010/10/01 9:39:10')
,(7, 101,'Start'  , '2010/10/01 9:39:10')
,(8, 102,'Start'  , '2010/10/01 9:39:10')
,(9, 101,'End'    , '2010/10/01 9:55:10');

<强>查询：

SELECT D.session_id
    ,TIMESTAMPDIFF(MINUTE,MIN(D.created), MAX(D.created)) time_count
FROM(
  SELECT a.id,a.session_id,a.created
    ,ROUND(count(*)/2,0) as RN 
    ,count(*) as row_number1 
  FROM a AS a
  JOIN a AS b ON a.session_id = b.session_id AND a.id >= b.id
  GROUP BY a.id,a.session_id,a.created
  ORDER BY 2,3
    )D
GROUP BY D.session_id,D.RN
HAVING COUNT(1)>1

<强>输出：

session_id  time_count
101         2
102         7
103         4
101         16

<强>小提琴：

检查 #SQL Fiddle

中的输出

Answer 3

你可以尝试一下sintax：

    WITH cte AS (
    SELECT
        session_id,
        DATEDIFF(minute, created, LAG(created, 1)) AS diff,
        ROW_NUMBER() OVER (PARTITION BY session_id ORDER BY created) rn
    FROM yourTable
)

SELECT

        session_id,
        diff AS time_count
    FROM cte
    WHERE rn % 2 = 0

;

如何计算组数据的日期差异

3 个答案: