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我正在尝试使用Haversine 公式从 WordPress 数据库中获取最近的位置

我的表结构如下

帖子

+--------------+
| Field        |
+--------------+
| ID           |
| post_author  |
| post_title   | 
| post_type    | 
+--------------+

后元

+--------------+
| Field        |
+--------------+
| meta_id      |
| post_id      |
| meta_key     |
| meta_value   |
+--------------+

并有记录meta_key值latitude和longitude

有关我用来获取纬度和经度的 SQL，请参阅我上一个问题的答案。

SELECT p.ID, 
  p.post_title, 
  p.post_author,
  max(case when pm.meta_key='latitude' then pm.meta_value end) latitude,
  max(case when pm.meta_key='longitude' then pm.meta_value end) longitude
FROM `wp_posts` p
LEFT JOIN `wp_postmeta` pm
  on p.ID=pm.post_id 
WHERE p.post_type='place' 
  AND (pm.meta_key='latitude' OR pm.meta_key='longitude') 
GROUP BY p.ID, p.post_title, p.post_author
ORDER BY p.ID ASC

现在我想将上面的查询合并到这个问题的答案中

SELECT item1, item2, 
    ( 3959 * acos( cos( radians(37) ) 
                   * cos( radians( lat ) ) 
                   * cos( radians( lng ) 
                       - radians(-122) ) 
                   + sin( radians(37) ) 
                   * sin( radians( lat ) ) 
                 )
   ) AS distance 
FROM geocodeTable 
HAVING distance < 25 
ORDER BY distance LIMIT 0 , 20;

以下是通过组合查询

SELECT ID, 
  post_title, 
  post_author,
  max(case when meta_key='latitude' then meta_value end) latitude,
  max(case when meta_key='longitude' then meta_value end) longitude,
  ( 3959 * acos( cos( radians(18.204540500000) ) 
                   * cos( radians( latitude ) ) 
                   * cos( radians( longitude ) 
                       - radians(-66.450958500000) ) 
                   + sin( radians(18.204540500000 ) 
                   * sin( radians( latitude ) ) 
                 )
   ) AS distance 
FROM `wp_posts` 
LEFT JOIN `wp_postmeta` 
  on ID=post_id 
WHERE post_type='place' 
  AND (meta_key='latitude' OR meta_key='longitude') 
GROUP BY ID, post_title, post_author
ORDER BY ID ASC

但这会产生语法错误

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'AS distance FROM `wp_posts` LEFT JOIN `wp_postmeta` on ID=post_id WHERE po' at line 13
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2 回答 2

4

你错过了)第一个结束sin()

( 3959 * acos( cos( radians(18.204540500000) ) 
                   * cos( radians( latitude ) ) 
                   * cos( radians( longitude ) 
                       - radians(-66.450958500000) ) 
                   + sin( radians(18.204540500000 ) ) /* <--- here */
                   * sin( radians( latitude ) ) 
              )
 ) AS distance

虽然很难直观地发现，但我通过将您的代码复制到支持大括号匹配的文本编辑器中找到了这一点。强烈建议使用一个，如果不是用于查询开发和测试，那么至少用于调试。

于 2012-11-06T12:09:22.803 回答
0

尝试这个：

SELECT *, ( 3959 * ACOS( COS( RADIANS(18.204540500000) ) 
                   * COS( RADIANS( latitude ) ) 
                   * COS( RADIANS( longitude ) 
                       - RADIANS(-66.450958500000) ) 
                   + SIN( RADIANS(18.204540500000 ) )
                   * SIN( RADIANS( latitude ) ) 
                 )
   ) AS distance  
FROM 
(SELECT ID, 
  post_title, 
  post_author,
  MAX(CASE WHEN meta_key='geo_latitude' THEN meta_value END) latitude,
  MAX(CASE WHEN meta_key='geo_longitude' THEN meta_value END) longitude
FROM `wp_posts` 
LEFT JOIN `wp_postmeta` 
  ON ID=post_id 
WHERE post_type='place' 
  AND (meta_key='geo_latitude' OR meta_key='geo_longitude') 
GROUP BY ID, post_title, post_author
ORDER BY ID ASC) AS A
于 2012-11-06T12:53:19.960 回答