doujiejujixi27244 2011-12-02 05:56
浏览 42
已采纳

MYSQL / PHP - 返回行的平均值而值不变?

The title of this doesn't quite make sense, so I'll do my best to explain.

I have a very large dataset (1000's of rows) in a single table. The data in this table relates to GPS tracking of vehicles. When the vehicle is stationary (Speed=0), the latitude and longitude can vary quite dramatically over a period of 12 hours.

My current SELECT query is this:

$query = "SELECT UUID, UNITID, Truncate(LONGITUDE,6) AS LONGITUDE, Truncate(LATITUDE,6) AS LATITUDE, SPEED, TRACKINGTIME FROM trackpoint_au WHERE SPEED > -1 Order By UnitID, TRACKINGTIME";

The query returns an XML page, via PHP. Built like so:

header("Content-type: text/xml");
// Start XML file, echo parent node
echo '<markers>';
// Iterate through the rows, printing XML nodes for each
while ($row = @mysql_fetch_assoc($result)){ 
// ADD TO XML DOCUMENT NODE
echo '<marker ';
echo 'unitid="' . $row['UNITID'] . '" ';
echo 'lat="' . $row['LATITUDE'] . '" ';
echo 'lng="' . $row['LONGITUDE'] . '" ';
echo 'spd="' . $row['SPEED'] . '" ';
echo 'time="' . $row['TRACKINGTIME'] . '" ';
echo '/>';
}
// End XML file
echo '</markers>';

The output looks like:

<marker unitid="7711010426" lat="-32.080402" lng="115.854890" spd="0" time="2011-11-30 06:15:00" />
<marker unitid="7711010426" lat="-32.080376" lng="115.854880" spd="0" time="2011-11-30 06:16:00" />
<marker unitid="7711010426" lat="-32.080364" lng="115.854880" spd="0" time="2011-11-30 06:17:00" />
<marker unitid="7711010426" lat="-32.080330" lng="115.854836" spd="0" time="2011-11-30 06:18:00" />
<marker unitid="7711010426" lat="-32.080326" lng="115.854860" spd="1.85" time="2011-11-30 06:20:00" />
<marker unitid="7711010426" lat="-32.080265" lng="115.854890" spd="0" time="2011-11-30 06:21:00" /> 
<marker unitid="7711010426" lat="-32.080276" lng="115.854920" spd="0" time="2011-11-30 06:22:00" /> 
<marker unitid="7711010426" lat="-32.080315" lng="115.854900" spd="0" time="2011-11-30 06:23:00" /> 
<marker unitid="7711010426" lat="-32.080296" lng="115.854866" spd="0" time="2011-11-30 06:24:00" />

My question is this: How can I use PHP OR MYSQL to return the average latitude/longitude of the rows with spd=0?

My resulting data should be like this:

<marker unitid="7711010426" lat="-32.080367" lng="115.8548715" spd="0" time="2011-11-30 06:18:00" />
<marker unitid="7711010426" lat="-32.080326" lng="115.854860" spd="1.85" time="2011-11-30 06:20:00" />
<marker unitid="7711010426" lat="-32.080288" lng="115.854894" spd="0" time="2011-11-30 06:24:00" />

Note, the 'average' row has the LAST timestamp of the rows that have been averaged out.

I have tried to use a 'Group By Speed'. However, this fails to do what I need as it groups ALL the records with an identical speed, not just the ones with a value of 0.

EDIT

Grouping by the UUID as suggested by macek does not help, as the UUID is unique for each row.

<marker time="2011-11-30 06:15:00" spd="0" lng="115.854890" lat="-32.080402" unitid="7711010426" uuid="c6d50454-aa5b-4069-8756-72c787923173"/>
<marker time="2011-11-30 06:16:00" spd="0" lng="115.854880" lat="-32.080376" unitid="7711010426" uuid="be6f9052-ab00-430a-8cec-6abf5051cad1"/>

ANSWER

After posting the question and reading some of the answers below, I managed to put this PHP code together. It loops through all the rows, checks the speed, if the speed is 0, check the next row (until speed<>0) and average out the lat/lng of those points.

for($i=0;$i<$num;$i++){
    mysql_data_seek($result,$i); 
    $row = mysql_fetch_assoc($result); 
    if ($row['SPEED']==0){
    //echo $i . ' spd: '.$row['SPEED'] . '<br />';
    $spd0 = true;
    $counter = 1;
    $lat = $row['LATITUDE'];
    $lng = $row['LONGITUDE'];
    $i++;
    while (($spd0==true) && ($i<$num)){
        //echo ' + ' . $i;
        mysql_data_seek($result,$i); 
        $row2 = mysql_fetch_assoc($result);
        if (($row2['UNITID']==$row['UNITID']) && ($row2['SPEED']==0)){
            $counter++;
            $lat = $lat + $row2['LATITUDE'];
            $lng = $lng + $row2['LONGITUDE'];
            //echo $i . ' spd: '.$row2['SPEED'] . '<br />';
            $i++;
        }
        else{
            $spd0=false;
            $i--;
        }
    }
    $lat = $lat/$counter;
    $lng = $lng/$counter;

    // ADD TO XML DOCUMENT NODE
    echo '<marker ';
    echo 'uuid ="' . $row['UUID'] . '" ';
    echo 'unitid="' . $row['UNITID'] . '" ';
    echo 'lat="' . $lat . '" ';
    echo 'lng="' . $lng . '" ';
    echo 'spd="' . $row['SPEED'] . '" ';
    echo 'time="' . $row['TRACKINGTIME'] . '" ';
    echo '/>';
}
else {
    //echo $i;
    // ADD TO XML DOCUMENT NODE
    echo '<marker ';
    echo 'uuid ="' . $row['UUID'] . '" ';
    echo 'unitid="' . $row['UNITID'] . '" ';
    echo 'lat="' . $row['LATITUDE'] . '" ';
    echo 'lng="' . $row['LONGITUDE'] . '" ';
    echo 'spd="' . $row['SPEED'] . '" ';
    echo 'time="' . $row['TRACKINGTIME'] . '" ';
    echo '/>';
}
} 

If someone has a more elegant way of checking the next rows, please post it, as always looking for ways to improve my code.

Thanks all!

  • 写回答

4条回答 默认 最新

  • dongqiya9552 2011-12-02 08:01
    关注

    in addition to the normal use of GROUP BY and AVG() you may be interested in Quassnoi's answer to my question here:

    GROUP BY for continuous rows in SQL

    He posted a very nice solution that also performs very well with many rows.

    Think of the speed as a state, and you want to aggregate all continous rows within a time period that have the same speed.

    Here is my attempt on rewriting your query using this method:

    SELECT 
            UNITID,
            /* we aggregate multiple rows, maybe you want to know which ones..
               this one is optional */
            CAST(GROUP_CONCAT(UUID SEPARATOR ', ') AS CHAR) AS UUIDS, 
            /* is group field in the inner subquery, we can just use it 
               in our select without an aggregate function */
            SPEED, 
            /* very important to select the lowest timestamp - 
               this is the time when your unit has stopped moving ..
               first row with speed=0 */
            MIN(TRACKINGTIME) AS TRACKINGTIME, 
            /* we calc the average on latitude here */
            TRUNCATE(AVG(LATITUDE),6) AS LATITUDE, 
            /* same for longitude */
            TRUNCATE(AVG(LONGITUDE),6) AS LONGITUDE, 
            /* maybe you want to know how many rows with speed 0 
               are grouped together */
            COUNT(UUID) AS AGGREGATE_COUNT 
    
    FROM    (
            SELECT
                    /* this increases the counter variable @r each time
                       the state has changed.. when speed of the previous row
                       was also "0" and is "0" in the current row, 
                       the counter is not increased. -- this is a virtual field 
                       we will use for GROUPing.
    
                       @speed is used to remember the speed of the previous
                       row for comparison in @r to determine if the speed has changed
                    */
                    @r := @r + (@prev_unit != UNITID 
                                  OR @prev_speed != 0 
                                  OR SPEED != 0) AS gn,  
                    @prev_speed := SPEED AS a_speed,
                    @prev_unit := UNITID AS a_unit,
                    tp.*
            FROM    (
                    SELECT  @r := 0,
                            @prev_speed := 1,
                            @prev_unit := ''
                    ) vars,
                    trackpoint_au tp
            ORDER BY
                    UNITID, TRACKINGTIME
            ) q
    GROUP BY
            gn
    ORDER BY
            UNITID
    

    Test data:

    CREATE TABLE `trackpoint_au` (
     `uuid` int(11) NOT NULL AUTO_INCREMENT,
     `latitude` decimal(10,0) NOT NULL,
     `longitude` decimal(10,0) NOT NULL,
     `speed` int(11) NOT NULL,
     `unitid` int(11) NOT NULL,
     `trackingtime` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
     PRIMARY KEY (`uuid`)
    ) ENGINE=MyISAM;
    
    INSERT INTO trackpoint_au (unitid, speed, latitude, longitude, trackingtime) VALUES
    (1, 0, 10, 10, NOW()),
    (1, 0, 20, 20, NOW()),
    (1, 1, 10, 10, NOW()),
    (1, 0, 10, 10, NOW()),
    (1, 0, 30, 30, NOW()),
    (2, 0, 10, 10, NOW()),
    (2, 0, 20, 20, NOW()),
    (3, 1, 10, 10, NOW()),
    (4, 0, 10, 10, NOW()),
    (4, 0, 20, 20, NOW()),
    (4, 1, 30, 30, NOW()),
    (4, 0, 60, 60, NOW()),
    (4, 0, 60, 60, NOW());
    

    Result:

    +--------+--------+-------+---------------------+-----------+-----------+-----------------+
    | UNITID | UUIDS  | SPEED | TRACKINGTIME        | LATITUDE  | LONGITUDE | AGGREGATE_COUNT |
    +--------+--------+-------+---------------------+-----------+-----------+-----------------+
    |      1 | 2, 1   |     0 | 2011-12-05 09:34:13 | 15.000000 | 15.000000 |               2 |
    |      1 | 3      |     1 | 2011-12-05 09:34:13 | 10.000000 | 10.000000 |               1 |
    |      1 | 4, 5   |     0 | 2011-12-05 09:34:13 | 20.000000 | 20.000000 |               2 |
    |      2 | 6, 7   |     0 | 2011-12-05 09:34:13 | 15.000000 | 15.000000 |               2 |
    |      3 | 8      |     1 | 2011-12-05 09:34:13 | 10.000000 | 10.000000 |               1 |
    |      4 | 9, 10  |     0 | 2011-12-05 09:34:13 | 15.000000 | 15.000000 |               2 |
    |      4 | 11     |     1 | 2011-12-05 09:34:13 | 30.000000 | 30.000000 |               1 |
    |      4 | 12, 13 |     0 | 2011-12-05 09:34:13 | 60.000000 | 60.000000 |               2 |
    +--------+--------+-------+---------------------+-----------+-----------+-----------------+
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(3条)

报告相同问题?

悬赏问题

  • ¥15 ubuntu虚拟机打包apk错误
  • ¥199 rust编程架构设计的方案 有偿
  • ¥15 回答4f系统的像差计算
  • ¥15 java如何提取出pdf里的文字?
  • ¥100 求三轴之间相互配合画圆以及直线的算法
  • ¥100 c语言,请帮蒟蒻写一个题的范例作参考
  • ¥15 名为“Product”的列已属于此 DataTable
  • ¥15 安卓adb backup备份应用数据失败
  • ¥15 eclipse运行项目时遇到的问题
  • ¥15 关于#c##的问题:最近需要用CAT工具Trados进行一些开发