我正在创建一个产品页面，买家只能从附近的商店根据卖家商店坐标和买家坐标订购。一切都很好，除了搜索功能，当我搜索产品时，它会显示数据库中存储的所有项目，但我不想这样，我还想使用mysql的搜索查询距离。
在mysql中，我使用union搜索卖家和产品。我的问题是如何实现这两个mysql UNION, INNER JOIN 以及 DISTANCE 一起。下面是我的表格结构。
产品

id   | podkey  | price  | name     | store_key  | owner_pkey  | pod_status
-----|---------|--------|----------|------------|-------------|------------
 1   | 1ab2    | 100    | Dog      | sk76n      | MBYUU00     | 1
 2   | 0kb2    | 180    | Cat      | sk76n      | MBYUU00     | 1
 3   | 5ab1    | 101    | Doggy    | sk76n      | MBYUU00     | 1
 4   | maf1    | 60     | Bird     | sk16b      | MBYUU00     | 1
 5   | baf0    | 90     | Dog      | sk78x      | KLYUP07     | 1

商店

id  | skey  | ownerkey | lat        | lng        | name
----|-------|----------|------------|------------|----------------- 
1   | sk76n | MBYUU00  | 2.922597   | 101.661896 | Cyberjaya store
2   | sk16b | MBYUU00  | 3.7384847  | 107.893837 | Puchong store
3   | sk78x | KLYUP07  | 7.64874646 | 883.893839 | My only store

业主

id  | name    | owner_key 
----|---------|--------------
1   | peter   | MBYUU00
2   | john    | KLYUP07

下面这是什么是我目前正在使用的显示页面上的项目，并正在工作

SELECT p.id, p.podkey, p.price, p.name, p.store_key, p.pod_status, s.id, s.skey, s.ownerkey, s.name, o.id, o.name, o.owner_key,
SQRT(POW(69.1 * (s.lat - :getBuyerLat), 2) + POW(69.1 * (:getBuyerLon - s.lng) * COS(s.lat / 57.3), 2)) AS distance
FROM product p

INNER JOIN Owners o
ON p.owner_pkey = o.owner_key

INNER JOIN store s
ON s.skey = p.store_key

WHERE p.pod_status = 1
HAVING distance <= 5

下面的查询是我在搜索中尝试的，但它没有返回任何内容或显示任何错误

SELECT * FROM ( 
    SELECT 'products' as type, pod_status AS stat, name AS name, price AS price, store_key AS skey, NULL AS distance FROM product p
    UNION
    SELECT 'sellers' as type, 1 AS stat, name AS name, NULL AS price, NULL AS skey, NULL AS distance FROM Owners o
    UNION
    SELECT 'outlets' as type, 1 AS stat, name AS name, NULL AS price, skey AS skey,
    SQRT(POW(69.1 * (s.lat - :getBuyerLat), 2) + POW(69.1 * (:getBuyerLon - s.lng) * COS(s.lat / 57.3), 2)) AS distance
    FROM store s
) sfoo
INNER JOIN store sj
ON sj.skey = sfoo.store_key
WHERE sfoo.name LIKE %dog% AND sfoo.stat = 1 HAVING distance <= 5

有人能帮我吗？我是麦克维http://sqlfiddle.com/#!9/AF39/2/0号