如何找到最近的定位点？
How to find the nearest location points?
我有一个 table，其中包含 customer_number、customer_location（纬度、经度）、current_latched_tower_ID、tower_location（纬度、经度）和距离（在 cx 位置和锁定塔位置之间）。我正在尝试找到离客户位置更近的塔，而不是 current_latched_tower。
示例：一位客户目前锁定了一个塔（KA001），客户位置与锁定的塔位置之间的距离为 1.6KM。然而，还有另一个塔（KA002）离客户位置很近。因此，其客户位置与最近的塔位置之间的距离为1.3KM。
Table
customer_number cx_lat cx_long tower_lat tower_long Latched_tower_ID Distance
34532 6.897257333 79.86474533 6.890487 79.869199 CM0321 0.51477
43445 6.935598403 81.14939421 6.947618 81.160246 BD0010 1.2292
54365 6.866224 79.88215 6.896111 79.868611 CM0037 1.6216
52568 7.113198 80.037247 7.121666 80.028888 GM0121 0.9476
期待输出table
customer_number cx_lat cx_long tower_lat tower_long Latched_tower_ID Distance Cloesed_tower_ID Closed_Distance
34532 6.897257333 79.86474533 6.890487 79.869199 CM0321 0.51477 CM0037 0.43222
52568 7.113198 80.037247 7.121666 80.028888 GM0121 0.9476 NULL NULL
如果没有任何更近的塔而不是锁定塔“Cloesed_tower_ID”和“Closed_Distance”列应该为 NULL
如果您不使用 SDO_GEOM，那么您可以创建半正弦函数来计算两个 latitude/longditude 坐标之间的距离：
CREATE FUNCTION haversine_distance(
lat1 IN NUMBER,
long1 IN NUMBER,
lat2 IN NUMBER,
long2 IN NUMBER
) RETURN NUMBER DETERMINISTIC
IS
PI CONSTANT NUMBER := ASIN(1) * 2;
R CONSTANT NUMBER := 6371000; -- Approx. radius of the earth in m
PHI1 CONSTANT NUMBER := lat1 * PI / 180;
PHI2 CONSTANT NUMBER := lat2 * PI / 180;
DELTA_PHI CONSTANT NUMBER := (lat2 - lat1) * PI / 180;
DELTA_LAMBDA CONSTANT NUMBER := (long2 - long1) * PI / 180;
a NUMBER;
c NUMBER;
BEGIN
a := SIN(delta_phi/2) * SIN(delta_phi/2) + COS(phi1) * COS(phi2) *
SIN(delta_lambda/2) * SIN(delta_lambda/2);
c := 2 * ATAN2(SQRT(a), SQRT(1-a));
RETURN R * c; -- in metres
END;
/
然后，我假设您的数据为第三范式：
CREATE TABLE towers (
tower_id PRIMARY KEY,
t_lat,
t_long
) AS
SELECT 'CM0321', 6.890487, 79.869199 FROM DUAL UNION ALL
SELECT 'BD0010', 6.947618, 81.160246 FROM DUAL UNION ALL
SELECT 'CM0037', 6.896111, 79.868611 FROM DUAL UNION ALL
SELECT 'GM0121', 7.121666, 80.028888 FROM DUAL;
CREATE TABLE customers (
customer_number PRIMARY KEY,
cx_lat,
cx_long,
latched_tower_id
) AS
SELECT 34532, 6.897257333, 79.86474533, 'CM0321' FROM DUAL UNION ALL
SELECT 43445, 6.935598403, 81.14939421, 'BD0010' FROM DUAL UNION ALL
SELECT 54365, 6.866224000, 79.88215000, 'CM0037' FROM DUAL UNION ALL
SELECT 52568, 7.113198000, 80.03724700, 'GM0121' FROM DUAL;
ALTER TABLE customers ADD CONSTRAINT customers__lti__fk
FOREIGN KEY (latched_tower_id) REFERENCES towers (tower_id);
然后，从 Oracle 12 开始，您可以使用以下方法计算更近的塔：
SELECT c.*,
TO_CHAR(
HAVERSINE_DISTANCE(c.cx_lat, c.cx_long, t.t_lat, t.t_long)/1000,
'FM999990.000'
) AS distance,
ct.tower_id AS closer_tower_id,
TO_CHAR(ct.distance, 'FM999990.000') AS closer_distance
FROM customers c
INNER JOIN towers t
ON (t.tower_id = c.latched_tower_id)
LEFT OUTER JOIN LATERAL(
SELECT ct.*,
HAVERSINE_DISTANCE(
c.cx_lat,
c.cx_long,
ct.t_lat,
ct.t_long
)/1000 AS distance
FROM towers ct
ORDER BY distance ASC
FETCH FIRST ROW ONLY
) ct
ON (ct.tower_id != c.latched_tower_id);
输出：
CUSTOMER_NUMBER
CX_LAT
CX_LONG
LATCHED_TOWER_ID
DISTANCE
CLOSER_TOWER_ID
CLOSER_DISTANCE
43445
6.935598403
81.14939421
BD0010
1.795
54365
6.866224
79.88215
CM0037
3.644
CM0321
3.053
34532
6.897257333
79.86474533
CM0321
0.899
CM0037
0.445
52568
7.113198
80.037247
GM0121
1.318
Oracle 12之前，您可以使用：
SELECT customer_number,
cx_lat,
cx_long,
latched_tower_id,
distance,
CASE
WHEN latched_tower_id != closer_tower_id
THEN closer_tower_id
END AS closer_tower_id,
CASE
WHEN latched_tower_id != closer_tower_id
THEN closer_distance
END AS closer_distance
FROM (
SELECT c.*,
TO_CHAR(
HAVERSINE_DISTANCE(c.cx_lat, c.cx_long, t.t_lat, t.t_long)/1000,
'FM999990.000'
) AS distance,
ct.tower_id AS closer_tower_id,
TO_CHAR(
HAVERSINE_DISTANCE(c.cx_lat, c.cx_long, ct.t_lat, ct.t_long)/1000,
'FM999990.000'
) AS closer_distance,
ROW_NUMBER() OVER (
PARTITION BY c.customer_number
ORDER BY HAVERSINE_DISTANCE(c.cx_lat, c.cx_long, ct.t_lat, ct.t_long)
) AS rn
FROM customers c
INNER JOIN towers t
ON (t.tower_id = c.latched_tower_id)
CROSS JOIN towers ct
ORDER BY
customer_number,
DISTANCE ASC
)
WHERE rn = 1;
db<>fiddle here
我有一个 table，其中包含 customer_number、customer_location（纬度、经度）、current_latched_tower_ID、tower_location（纬度、经度）和距离（在 cx 位置和锁定塔位置之间）。我正在尝试找到离客户位置更近的塔，而不是 current_latched_tower。
示例：一位客户目前锁定了一个塔（KA001），客户位置与锁定的塔位置之间的距离为 1.6KM。然而，还有另一个塔（KA002）离客户位置很近。因此，其客户位置与最近的塔位置之间的距离为1.3KM。
Table
customer_number cx_lat cx_long tower_lat tower_long Latched_tower_ID Distance
34532 6.897257333 79.86474533 6.890487 79.869199 CM0321 0.51477
43445 6.935598403 81.14939421 6.947618 81.160246 BD0010 1.2292
54365 6.866224 79.88215 6.896111 79.868611 CM0037 1.6216
52568 7.113198 80.037247 7.121666 80.028888 GM0121 0.9476
期待输出table
customer_number cx_lat cx_long tower_lat tower_long Latched_tower_ID Distance Cloesed_tower_ID Closed_Distance
34532 6.897257333 79.86474533 6.890487 79.869199 CM0321 0.51477 CM0037 0.43222
52568 7.113198 80.037247 7.121666 80.028888 GM0121 0.9476 NULL NULL
如果没有任何更近的塔而不是锁定塔“Cloesed_tower_ID”和“Closed_Distance”列应该为 NULL
如果您不使用 SDO_GEOM，那么您可以创建半正弦函数来计算两个 latitude/longditude 坐标之间的距离：
CREATE FUNCTION haversine_distance(
lat1 IN NUMBER,
long1 IN NUMBER,
lat2 IN NUMBER,
long2 IN NUMBER
) RETURN NUMBER DETERMINISTIC
IS
PI CONSTANT NUMBER := ASIN(1) * 2;
R CONSTANT NUMBER := 6371000; -- Approx. radius of the earth in m
PHI1 CONSTANT NUMBER := lat1 * PI / 180;
PHI2 CONSTANT NUMBER := lat2 * PI / 180;
DELTA_PHI CONSTANT NUMBER := (lat2 - lat1) * PI / 180;
DELTA_LAMBDA CONSTANT NUMBER := (long2 - long1) * PI / 180;
a NUMBER;
c NUMBER;
BEGIN
a := SIN(delta_phi/2) * SIN(delta_phi/2) + COS(phi1) * COS(phi2) *
SIN(delta_lambda/2) * SIN(delta_lambda/2);
c := 2 * ATAN2(SQRT(a), SQRT(1-a));
RETURN R * c; -- in metres
END;
/
然后，我假设您的数据为第三范式：
CREATE TABLE towers (
tower_id PRIMARY KEY,
t_lat,
t_long
) AS
SELECT 'CM0321', 6.890487, 79.869199 FROM DUAL UNION ALL
SELECT 'BD0010', 6.947618, 81.160246 FROM DUAL UNION ALL
SELECT 'CM0037', 6.896111, 79.868611 FROM DUAL UNION ALL
SELECT 'GM0121', 7.121666, 80.028888 FROM DUAL;
CREATE TABLE customers (
customer_number PRIMARY KEY,
cx_lat,
cx_long,
latched_tower_id
) AS
SELECT 34532, 6.897257333, 79.86474533, 'CM0321' FROM DUAL UNION ALL
SELECT 43445, 6.935598403, 81.14939421, 'BD0010' FROM DUAL UNION ALL
SELECT 54365, 6.866224000, 79.88215000, 'CM0037' FROM DUAL UNION ALL
SELECT 52568, 7.113198000, 80.03724700, 'GM0121' FROM DUAL;
ALTER TABLE customers ADD CONSTRAINT customers__lti__fk
FOREIGN KEY (latched_tower_id) REFERENCES towers (tower_id);
然后，从 Oracle 12 开始，您可以使用以下方法计算更近的塔：
SELECT c.*,
TO_CHAR(
HAVERSINE_DISTANCE(c.cx_lat, c.cx_long, t.t_lat, t.t_long)/1000,
'FM999990.000'
) AS distance,
ct.tower_id AS closer_tower_id,
TO_CHAR(ct.distance, 'FM999990.000') AS closer_distance
FROM customers c
INNER JOIN towers t
ON (t.tower_id = c.latched_tower_id)
LEFT OUTER JOIN LATERAL(
SELECT ct.*,
HAVERSINE_DISTANCE(
c.cx_lat,
c.cx_long,
ct.t_lat,
ct.t_long
)/1000 AS distance
FROM towers ct
ORDER BY distance ASC
FETCH FIRST ROW ONLY
) ct
ON (ct.tower_id != c.latched_tower_id);
输出：
CUSTOMER_NUMBER CX_LAT CX_LONG LATCHED_TOWER_ID DISTANCE CLOSER_TOWER_ID CLOSER_DISTANCE 43445 6.935598403 81.14939421 BD0010 1.795 54365 6.866224 79.88215 CM0037 3.644 CM0321 3.053 34532 6.897257333 79.86474533 CM0321 0.899 CM0037 0.445 52568 7.113198 80.037247 GM0121 1.318
Oracle 12之前，您可以使用：
SELECT customer_number,
cx_lat,
cx_long,
latched_tower_id,
distance,
CASE
WHEN latched_tower_id != closer_tower_id
THEN closer_tower_id
END AS closer_tower_id,
CASE
WHEN latched_tower_id != closer_tower_id
THEN closer_distance
END AS closer_distance
FROM (
SELECT c.*,
TO_CHAR(
HAVERSINE_DISTANCE(c.cx_lat, c.cx_long, t.t_lat, t.t_long)/1000,
'FM999990.000'
) AS distance,
ct.tower_id AS closer_tower_id,
TO_CHAR(
HAVERSINE_DISTANCE(c.cx_lat, c.cx_long, ct.t_lat, ct.t_long)/1000,
'FM999990.000'
) AS closer_distance,
ROW_NUMBER() OVER (
PARTITION BY c.customer_number
ORDER BY HAVERSINE_DISTANCE(c.cx_lat, c.cx_long, ct.t_lat, ct.t_long)
) AS rn
FROM customers c
INNER JOIN towers t
ON (t.tower_id = c.latched_tower_id)
CROSS JOIN towers ct
ORDER BY
customer_number,
DISTANCE ASC
)
WHERE rn = 1;
db<>fiddle here