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oracle - 如何找到最近的定位点？

问题描述

我有一张表，其中包含 customer_number、customer_location(lat, long)、current_latched_tower_ID、tower_location(lat,long) 和距离（在 cx 位置和锁定塔位置之间）。我试图找到一个离客户位置非常近的塔，而不是 current_latched_tower。

示例：一位客户当前锁定一个塔（KA001），客户位置与锁定塔位置之间的距离为1.6KM。但是，还有另一个塔（KA002）非常靠近客户位置。因此，其客户位置与最近的塔位置之间的距离为 1.3KM。

桌子

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

期望输出表

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

如果没有更近的塔而不是锁定塔“Closed_tower_ID”和“Closed_Distance”列应为 NULL

标签： oracleoracle11g

解决方案

如果您不使用，SDO_GEOM那么您可以创建 hasrsine 函数来计算两个纬度/经度坐标之间的距离：

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，我假设您的数据采用第三范式：

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，在 Oracle 12 中，您可以使用以下方法计算更接近的塔：

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);

哪个输出：

顾客号码 CX_LAT CX_LONG LATCHED_TOWER_ID 距离 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

在 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<>在这里摆弄

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