Question

我有一张包含位置和时间信息的表gps（时间，纬度，长度）。我有一个表事件，可以保留某些事件（时间）的时间。

我需要sql（最好是mysql）创建一个新表：header('Content-Type: application/json'); $arr=array("Download" => "http://www.exemple.com/file"); $json = json_encode($arr); print_r($json);对于表事件中的每个记录，注册表events_with_gps的寄存器的位置（lat，long）更接近事件。

示例：

gps

而且，是否可以将gps: 11:00:00, 10, 10 11:05:00, 14, 14 11:10:00, 18, 18 11:15:00, 20, 20 11:20:00, 22, 22 events: 11:06:02 11:14:09 events_with_gps: 11:06:02, 14, 14 11:14:09, 20, 20中的位置设置为上一个和下一个gps寄存器之间的平均值？

events_with_gps

Answer 1

在最近时间执行联接的一种方法是首先使用CROSS JOIN来计算gps和event表格中时间列之间的所有可能差异。

以下查询：

SELECT e.`time` AS etime, g.`time` AS gtime, g.lat, g.`long`, 
       ABS(TIMEDIFF(e.`time`, g.`time`)) AS diff
FROM events AS e
CROSS JOIN gps AS g
ORDER BY e.`time`, diff

产生以下输出：

etime       gtime       lat long    diff
------------------------------------------
11:06:02    11:05:00    14  14      102
11:06:02    11:10:00    18  18      358
11:06:02    11:00:00    10  10      602
11:06:02    11:15:00    20  20      858
11:06:02    11:20:00    22  22      1358
11:14:09    11:15:00    20  20      51
11:14:09    11:10:00    18  18      409
11:14:09    11:20:00    22  22      551
11:14:09    11:05:00    14  14      909
11:14:09    11:00:00    10  10      1409

如果您GROUP BY etime并使用GROUP_CONCAT，您可以从上述结果集中选择具有最小diff值的记录。您可以将JOIN这些记录返回gps，以便将记录插入events_with_gps：

SELECT t.`time`, g.lat, g.`long`
FROM (
SELECT e.`time`, 
       SUBSTRING_INDEX(
          GROUP_CONCAT(g.`time` 
                       ORDER BY ABS(TIMEDIFF(e.`time`, g.`time`))), ',', 1) AS matchedTime
FROM events AS e
CROSS JOIN gps AS g
GROUP BY e.`time` ) t
INNER JOIN gps AS g ON CAST(t.matchedTime AS TIME) = g.`time`

<强>输出：

 time       lat long
-----------------------
'11:06:02'  14  14
'11:14:09'  20  20

Demo here

要获取第二个结果集，您可以使用以下查询获取每个gps记录的上一个和下一个events记录：

SELECT e.`time` as etime, 
       (SELECT `time`
        FROM gps AS g
        WHERE g.`time` <= e.`time`
        ORDER BY g.`time` DESC 
        LIMIT 1) AS prevTime,
       (SELECT `time`
        FROM gps AS g
        WHERE g.`time` > e.`time`
        ORDER BY g.`time`  
        LIMIT 1) AS nextTime        
FROM events AS e

输出

etime     prevTime  nextTime
------------------------------
11:06:02  11:05:00  11:10:00
11:14:09  11:10:00  11:15:00

您现在可以将上述内容用作派生表，以便加入gps并获取“上一个”和“下一个”lat和{{1} }值：

long

<强>输出：

SELECT t.etime, (g1.lat + g2.lat) /2 , (g1.`long` + g2.`long`) / 2
FROM (
SELECT e.`time` as etime, 
       (SELECT `time`
        FROM gps AS g
        WHERE g.`time` <= e.`time`
        ORDER BY g.`time` DESC 
        LIMIT 1) AS prevTime,
       (SELECT `time`
        FROM gps AS g
        WHERE g.`time` > e.`time`
        ORDER BY g.`time`  
        LIMIT 1) AS nextTime        
FROM events AS e ) t
INNER JOIN gps AS g1 ON t.prevTime = g1.time
INNER JOIN gps AS g2 ON t.nextTime = g2.time

Demo here

SQL按最近时间合并2个表

1 个答案: