根據座標之間的距離選擇位置

Question

我嘗試製作一個代碼，用於計算某個給定座標的最近位置。然而，我陷入了以下錯誤：

ERROR: column "distance" does not exist 
LINE 5: `HAVING distance < 150` 
in ROOT\frontend3.php on line 16. 

數據從pgadminIII數據庫檢索。所有幫助是值得歡迎

<?php 

include 'connection.php'; 

$lat = $_GET['lat']; 
$lng = $_GET['lng']; 

#Select inormation from Parking database 
$result = pg_query($conn, " 
SELECT id, (6371 * acos(cos(radians($lat)) * cos(radians(lat)) *  cos(radians(lng) - radians($lng)) + sin(radians($lat)) * sin(radians(lat)))) 
AS distance 
FROM parking 
HAVING distance < 150 
ORDER BY distance 
LIMIT 10 
"); 
echo $result; 

Answer 1

having是當你組/彙總的東西，需要一個分組後的情況。你的情況，你需要使用where：用於聚合使用組時和數據集後應用

SELECT id, (...) AS distance 
FROM  parking 
WHERE distance < 150 
ORDER BY distance 
LIMIT 10 

Answer 2

SELECT 
    id 
    ,(6371 * acos(cos(radians($lat)) * cos(radians(lat)) *  cos(radians(lng) - radians($lng)) + sin(radians($lat)) * sin(radians(lat)))) AS distance 
FROM parking 
WHERE 
    (6371 * acos(cos(radians($lat)) * cos(radians(lat)) *  cos(radians(lng) - radians($lng)) + sin(radians($lat)) * sin(radians(lat)))) < 150 
ORDER BY distance 
LIMIT 10; 

2發佈具有使用。所以一個where語句會更適合你的。第二個距離是列別名，所以它不會在where子句中可用，通過將整個計算放在where子句中可以解決這個問題。或者使用子選擇。

SELECT * 
FROM (
    SELECT 
     id 
     ,(6371 * acos(cos(radians($lat)) * cos(radians(lat)) *  cos(radians(lng) - radians($lng)) + sin(radians($lat)) * sin(radians(lat)))) AS distance 
    FROM parking 
) t 
WHERE 
    distance < 150 
ORDER BY distance 
LIMIT 10;