求和每个mysql的结果

Question

表位：

+----+-----------+-----------+---------------------+
| Id | Old Value | New Value |        Date         |
+----+-----------+-----------+---------------------+
|  1 | Home      | Work      | 2016-11-01 09:00:00 |
|  2 | Work      | Home      | 2016-11-01 09:25:00 |
|  3 | Home      | Work      | 2016-11-01 09:44:59 |
|  4 | Work      | Home      | 2016-11-01 10:11:50 |
|  5 | Home      | Work      | 2016-11-01 10:25:30 |
|  6 | Work      | Home      | 2016-11-01 11:01:15 |
+----+-----------+-----------+---------------------+

我的查询：

    SELECT loc1.*,loc2.date AS loc2_date, SUM(CAST(TIMESTAMPDIFF(MICROSECOND,loc1.date,loc2.date)/1000000 AS DECIMAL(20))) AS time_diff
FROM locations AS loc1
        LEFT JOIN locations AS loc2 ON loc1.old_value=loc2.new_value AND loc2.id > loc1.id
        WHERE old_value="Home"

我的查询结果：

Id: 1
Old Value: Home
New Value: Work
Date: 2016-11-01 09:00:00
Loc2_date: 2016-11-01 09:25:00
Time_fidd: 1500

Id: 1
Old Value: Home
New Value: Work
Date: 2016-11-01 09:00:00
Loc2_date: 2016-11-01 10:11:50
Time_diff: 4310

Id: 1
Old Value: Home
New Value: Work
Date: 2016-11-01 09:00:00
Loc2_date: 2016-11-01 11:01:15
Time_diff: 7275

Id: 3
Old Value: Home
New Value: Work
Date: 2016-11-01 09:44:59
Loc2_date: 2016-11-01 10:11:50
Time_diff: 1611

Id: 3
Old Value: Home
New Value: Work
Date: 2016-11-01 09:44:59
Loc2_date: 2016-11-01 11:01:15
Time_diff: 4576

Id: 5
Old Value: Home
New Value: Work
Date: 2016-11-01 10:25:30
Loc2_date: 2016-11-01 11:01:15
Time_diff: 2145

我想总结一下日期，看看我上班的所有时间。如果我按ID分组将准确显示我需要的内容，但当我总结loc1.date和loc2.date之间的差异时，查询将跳过分组依次并继续返回所有行。请帮忙。

我的查询需要什么：

Id: 1
Old Value: Home
New Value: Work
Date: 2016-11-01 09:00:00
Loc2_date: 2016-11-01 09:25:00
Time_fidd: 1500

Id: 3
Old Value: Home
New Value: Work
Date: 2016-11-01 09:44:59
Loc2_date: 2016-11-01 10:11:50
Time_diff: 1611

Id: 5
Old Value: Home
New Value: Work
Date: 2016-11-01 10:25:30
Loc2_date: 2016-11-01 11:01:15
Time_diff: 2145

Answer 1

<强>更新

试试这个。

SELECT tablea.id, tablea.`Old_value`, tablea.`New_value`, tablea.`Date`, tablea.`Date` Loc2_date, SUM(tablea.time_diff) AS time_diff 
FROM 
(
    SELECT loc1.`id`, loc1.`Old_value`, loc1.`New_value`, loc1.`Date`, loc2.`Date` Loc2_date, CAST(ABS(TIMESTAMPDIFF(MICROSECOND, loc2.date, loc1.date)/1000000) AS DECIMAL(20)) time_diff
    FROM locations loc1 INNER JOIN locations loc2
    ON loc1.`Old_value`=loc2.`New_value` AND loc2.`id`>loc1.`id`
    WHERE loc1.`Old_value`="Home"
    GROUP BY loc1.`id`
) tablea    

示例：

SELECT tableb.*
FROM
(
    SELECT tablea.id, tablea.`Old_value`, tablea.`New_value`, tablea.`Date`, tablea.`Date` Loc2_date, SUM(tablea.time_diff) AS time_diff 
    FROM 
    (
        SELECT loc1.`id`, loc1.`Old_value`, loc1.`New_value`, loc1.`Date`, loc2.`Date` Loc2_date, CAST(ABS(TIMESTAMPDIFF(MICROSECOND, loc2.date, loc1.date)/1000000) AS DECIMAL(20)) time_diff
        FROM locations loc1 INNER JOIN locations loc2
        ON loc1.`Old_value`=loc2.`New_value` AND loc2.`id`>loc1.`id`
        WHERE loc1.`Old_value`="Home"
        GROUP BY loc1.`id`
    ) tablea    
) tableb

旧答案。

您可以尝试 group_by 。在GROUP BY loc1.id子句之后添加WHERE。

SELECT loc1.*,loc2.date AS loc2_date, SUM(CAST(TIMESTAMPDIFF(MICROSECOND,loc1.date,loc2.date)/1000000 AS DECIMAL(20)))
FROM locations AS loc1
LEFT JOIN locations AS loc2 ON loc1.old_value=loc2.new_value AND loc2.id > loc1.id
WHERE old_value="Home"
GROUP BY loc1.id