SQL元連接加Haversine公式

Question

我有一個查詢，它將一些元數據連接到用戶。

  SELECT 
       users.*, 
       gender.meta_value AS `gender`, 
       sexual_orientation.meta_value AS `sexual_orientation`, 
       relationship_status.meta_value AS `relationship_status`, 
       interest_1.meta_value AS `interest_1`, 
       interest_2.meta_value AS `interest_2`, 
       interest_3.meta_value AS `interest_3`, 
       interest_4.meta_value AS `interest_4`, 
       interest_5.meta_value AS `interest_5`, 
       interest_6.meta_value AS `interest_6`, 
       address.address_line_1, 
       address.address_line_2, 
       address.town, 
       address.county, 
       address.postcode, 
       address.country, 
       address.longitude, 
       address.latitude 
      FROM 
       `users` 
      JOIN 
       `storage_varchars` AS `gender` 
      ON 
       gender.user_id = users.id AND gender.meta_name = 'gender' 
      JOIN 
       `storage_varchars` AS `sexual_orientation` 
      ON 
       sexual_orientation.user_id = users.id AND sexual_orientation.meta_name = 'sexual_orientation' 
      JOIN 
       `storage_varchars` AS `relationship_status` 
      ON 
       relationship_status.user_id = users.id AND relationship_status.meta_name = 'relationship_status' 
      JOIN 
       `storage_varchars` AS `interest_1` 
      ON 
       interest_1.user_id = users.id AND interest_1.meta_name = 'interest_1' 
      JOIN 
       `storage_varchars` AS `interest_2` 
      ON 
       interest_2.user_id = users.id AND interest_2.meta_name = 'interest_2' 
      JOIN 
       `storage_varchars` AS `interest_3` 
      ON 
       interest_3.user_id = users.id AND interest_3.meta_name = 'interest_3' 
      JOIN 
       `storage_varchars` AS `interest_4` 
      ON 
       interest_4.user_id = users.id AND interest_4.meta_name = 'interest_4' 
      JOIN 
       `storage_varchars` AS `interest_5` 
      ON 
       interest_5.user_id = users.id AND interest_5.meta_name = 'interest_5' 
      JOIN 
       `storage_varchars` AS `interest_6` 
      ON 
       interest_6.user_id = users.id AND interest_6.meta_name = 'interest_6' 
      JOIN 
       `payments` AS `address` 
      ON 
       address.user_id = users.id 

我想現在使用它通過使用Haversine公式的最近距離搜索。

SELECT id, (3959 * acos(cos(radians(37)) * cos(radians(lat)) * cos(radians(lng) - radians(-122)) + sin(radians(37)) * sin(radians(lat)))) AS distance FROM markers HAVING distance < 25 ORDER BY distance LIMIT 0 , 20; 

我怎樣才能合併到兩個？

問候

Answer 1

好，我已經想出了一個辦法，但任何人都可以告訴我，如果這是正確的方法是什麼？

$usersLatitude = '53.1765254'; 
$usersLongitude = '-1.1954137'; 
$distance = 20; 

$query = " 
    SELECT main.* FROM 
    (
     SELECT 
      users.*, 
      gender.meta_value AS `gender`, 
      sexual_orientation.meta_value AS `sexual_orientation`, 
      relationship_status.meta_value AS `relationship_status`, 
      interest_1.meta_value AS `interest_1`, 
      interest_2.meta_value AS `interest_2`, 
      interest_3.meta_value AS `interest_3`, 
      interest_4.meta_value AS `interest_4`, 
      interest_5.meta_value AS `interest_5`, 
      interest_6.meta_value AS `interest_6`, 
      address.address_line_1, 
      address.address_line_2, 
      address.town, 
      address.county, 
      address.postcode, 
      address.country, 
      address.longitude, 
      address.latitude, 
      (3959 * acos(cos(radians({$usersLatitude})) * cos(radians(address.latitude)) * cos(radians(address.longitude) - radians({$usersLongitude})) + sin(radians({$usersLatitude})) * sin(radians(address.latitude)))) AS distance 
     FROM 
      `users` 
     JOIN 
      `storage_varchars` AS `gender` 
     ON 
      gender.user_id = users.id AND gender.meta_name = 'gender' 
     JOIN 
      `storage_varchars` AS `sexual_orientation` 
     ON 
      sexual_orientation.user_id = users.id AND sexual_orientation.meta_name = 'sexual_orientation' 
     JOIN 
      `storage_varchars` AS `relationship_status` 
     ON 
      relationship_status.user_id = users.id AND relationship_status.meta_name = 'relationship_status' 
     JOIN 
      `storage_varchars` AS `interest_1` 
     ON 
      interest_1.user_id = users.id AND interest_1.meta_name = 'interest_1' 
     JOIN 
      `storage_varchars` AS `interest_2` 
     ON 
      interest_2.user_id = users.id AND interest_2.meta_name = 'interest_2' 
     JOIN 
      `storage_varchars` AS `interest_3` 
     ON 
      interest_3.user_id = users.id AND interest_3.meta_name = 'interest_3' 
     JOIN 
      `storage_varchars` AS `interest_4` 
     ON 
      interest_4.user_id = users.id AND interest_4.meta_name = 'interest_4' 
     JOIN 
      `storage_varchars` AS `interest_5` 
     ON 
      interest_5.user_id = users.id AND interest_5.meta_name = 'interest_5' 
     JOIN 
      `storage_varchars` AS `interest_6` 
     ON 
      interest_6.user_id = users.id AND interest_6.meta_name = 'interest_6' 
     JOIN 
      `payments` AS `address` 
     ON 
      address.user_id = users.id 
    ) AS `main` 
    WHERE 
     `main`.distance < {$distance} 
    ORDER BY 
     `main`.distance 
";