mysql 计算返回位置的距离

Question

这不是以下内容的副本： Find distance between two points using latitude and longitude in mysql

I tried the query and it's not working, I get empty data set and 0 as the answer.

SEE answer, THAT is the solution, I don't know why this question was marked as duplicate from Shadow

我有一个 table 结构如下：

table: distance
id int(10) primary key
lat float,
lon float,
time timestamp

我有一个 android 应用程序可以 ping 我的服务器并在每次更改位置时添加一条记录。

例如

id:1, lat:40.753979, lon:-111.881721, time = 1571004620
id:2, lat:40.753979, lon:-111.883721, time = 1571004630
id:3, lat:40.753979, lon:-111.885721, time = 1571004640

如果我走一个方向，我可以计算出总距离：

/* in my python script*/
startLat,startLon = select lat,lon from distance where time >= 1571004620 order by time limit 1;
endLat,endLon = select lat,lon from distance where time <= 1571004640 order by time desc limit limit 1;

然后我可以减去两个坐标，最终得到 0.004000 的经度距离

问题：

如果我添加：

id:4, lat:40.753979, lon:-111.881721, time = 1571004650

那么我应该得到：0.008000

但我得到 0，因为位置 1 与位置 4 相同

Answer 1

您可以利用 MySQL 空间支持，从 5.7 开始可用，以在数据库中执行整个计算。函数 ST_Distance_Sphere() 可用于计算距离。

您实际上需要距离的累加和（这需要 window 函数，自 MySQL 8.0 起可用）。

考虑：

SELECT
    id,
    time,
    SUM(
        CASE WHEN lag_lat IS NULL 
            THEN 0
            ELSE ST_Distance_Sphere(point(lag_lon, lag_lat), point(lon, lat))
        END
    ) OVER (ORDER BY time) cumulative_distance
FROM (
    SELECT
        d.*,
        LAG(lat) OVER(ORDER BY time) lag_lat,
        LAG(lon) OVER(ORDER BY time) lag_lon
    FROM distance d
)  x

Demo on DB Fiddle:

| id  | time       | cumulative_distance |
| --- | ---------- | ------------------- |
| 1   | 1571004620 | 0                   |
| 2   | 1571004630 | 168.37177423236415  |
| 3   | 1571004640 | 336.7435484657999   |
| 4   | 1571004650 | 673.4870969097663   |

在 MySQL 的早期版本中，您需要模拟 window 函数：

LAG()可以替换为自身LEFT JOIN条件NOT EXISTS和ON条件[=22]中的相关子查询=]
变量可以模拟累积SUM

查询：

SELECT
    id,
    time,
    @running_distance := @running_distance + CASE
        WHEN lag_lat IS NULL THEN 0
        ELSE ST_Distance_Sphere(point(lag_lon, lag_lat), point(lon, lat))
    END running_distance
FROM (
    SELECT
        d.id,
        d.time,
        d.lat,
        d.lon,
        d_lag.lat lag_lat,
        d_lag.lon lag_lon   
    FROM distance d
    LEFT JOIN distance d_lag
        ON  d_lag.time < d.time
        AND NOT EXISTS (
            SELECT 1 
            FROM distance d1
            WHERE d1.time < d.time AND d1.time > d_lag.time
        )
    ORDER BY d.time
) x
CROSS JOIN (SELECT @running_distance := 0) y

Demo on DB Fiddle: 结果同上

mysql 计算返回位置的距离

mysql calculate distance for returnning location

python

mysql

mysql-spatial

mysql-5.7