Jadi, inilah solusi untuk bagian pertama dari masalah...
DROP TABLE IF EXISTS my_table;
CREATE TABLE my_table
(id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,user_name VARCHAR(12) NOT NULL
,date DATE NOT NULL
,LAT DECIMAL(5,3) NOT NULL
,LON DECIMAL (5,2) NOT NULL
);
INSERT INTO my_table VALUES
( 1,'maria','2005-01-01',51.555 ,5.014),
( 2,'maria','2005-01-01',51.437 ,5.474),
( 3,'peter','2005-02-03',51.437 ,5.474),
( 4,'john' ,'2005-02-03',51.858 ,5.864),
( 5,'maria','2005-02-04',51.858 ,5.864),
( 6,'john' ,'2005-02-03',51.437 ,5.474),
( 7,'john' ,'2006-02-04',0 ,0),
( 8,'john' ,'2006-02-04',51.858 ,5.864),
( 9,'john' ,'2006-02-04',51.858 ,5.864),
(10,'john' ,'2006-02-04',51.437 ,5.474);
SELECT x.user_name
, x.id from_id
, MIN(y.id) to_id
FROM my_table x
JOIN my_table y
ON y.user_name = x.user_name
AND y.id > x.id
WHERE (y.lat <> 0 AND y.lon <> 0)
AND (x.lat <> 0 AND x.lon <> 0)
GROUP
BY x.id;
+-----------+---------+-------+
| user_name | from_id | to_id |
+-----------+---------+-------+
| maria | 1 | 2 |
| maria | 2 | 5 |
| john | 4 | 6 |
| john | 6 | 8 |
| john | 8 | 9 |
| john | 9 | 10 |
+-----------+---------+-------+
Untuk sisa masalah, sesuatu seperti berikut ini akan berfungsi.
Saya memiliki fungsi di database saya yang disebut geo_distance_km. Ini terlihat seperti ini, dan menyimpan mengetikkan rumus haversine setiap kali:
delimiter //
create DEFINER = CURRENT_USER function geo_distance_km (lat1 double, lon1 double, lat2 double, lon2 double) returns double
begin
declare R int DEFAULT 6372.8;
declare phi1 double;
declare phi2 double;
declare d_phi double;
declare d_lambda double;
declare a double;
declare c double;
declare d double;
set phi1 = radians(lat1);
set phi2 = radians(lat2);
set d_phi = radians(lat2-lat1);
set d_lambda = radians(lon2-lon1);
set a = sin(d_phi/2) * sin(d_phi/2) +
cos(phi1) * cos(phi2) *
sin(d_lambda/2) * sin(d_lambda/2);
set c = 2 * atan2(sqrt(a), sqrt(1-a));
set d = R * c;
return d;
end;
//
delimiter ;
Kita bisa menggabungkannya dengan apa yang sudah kita miliki...
SELECT user_name
, YEAR(date) year
, COALESCE(SUM(distance),0) total
FROM
( SELECT a.*
, b.lat to_lat
, b.lon to_lon
, ROUND(geo_distance_km(from_lat,from_lon,b.lat,b.lon),3) distance
FROM
( SELECT x.user_name
, x.date
, x.id from_id
, x.lat from_lat
, x.lon from_lon
, MIN(y.id) to_id
FROM my_table x
LEFT
JOIN my_table y
ON y.user_name = x.user_name
AND y.id > x.id
AND (y.lat <> 0 OR y.lon <> 0)
WHERE (x.lat <> 0 AND x.lon <> 0)
GROUP
BY x.id
) a
LEFT
JOIN my_table b
ON b.id = a.to_id
) n
GROUP
BY user_name
, year;
+-----------+------+---------+
| user_name | year | total |
+-----------+------+---------+
| john | 2005 | 108.024 |
| john | 2006 | 54.012 |
| maria | 2005 | 88.464 |
| peter | 2005 | 0.000 |
+-----------+------+---------+
Saya tidak begitu mengerti bagaimana Anda menangani jarak yang tumpang tindih bertahun-tahun, tetapi ini akan membuat Anda mendekati apa yang Anda cari.
