Question

我有一个慈善表，其中包含字段：慈善机构，邮政编码和一个包含字段的邮政编码表：postcode，lat，lng

我想从网页发布邮政编码，找到最近的慈善机构

我是一个mysql初学者，所以我有点迷失，但我一直在尝试使用连接和子查询的各种想法，但没有一个工作（我要么得到语法错误，要么'操作数应该包含1列'以下代码的变化）我有

Select charity,postcode,
 ( 
   (Select lat as lat2, lng as lng2
    from postcodes
    where postcode='WN8'
    )

3959 * acos( cos( radians(lat2) ) * cos( radians( lat ) ) * 
cos( radians( lng ) - radians(lng2) ) + 
sin( radians(lat2) ) * sin( radians( lat ) ) ) 
  )
AS distance 
FROM postcodes  
JOIN   Charities on charities.postcode=postcodes.postcode
HAVING distance < 30 ORDER BY distance LIMIT 0 , 30;

我在这里看到很多例子，其中lat2和lng2是从发布的值中获得的，但不是从db中的表中获得的。

示例中的

p.s 'where postcode='WN8'仅用于测试

Answer 1

不确定您使用上述SQL获得了什么错误。

然而，尝试这个小调整，让我们知道你得到了什么错误

SELECT charity, postcode,
(3959 * acos( cos( radians(CustPostcode.lat) ) * cos( radians( postcodes.lat ) ) * 
cos( radians( postcodes.lng ) - radians(CustPostcode.lng) ) + 
sin( radians(CustPostcode.lat) ) * sin( radians( postcodes.lat ) ) ) 
  ) AS distance 
FROM postcodes  
INNER JOIN Charities ON charities.postcode=postcodes.postcode
CROSS JOIN  (SELECT lat, lng FROM postcodes WHERE postcode='WN8') CustPostcode
HAVING distance < 30 
ORDER BY distance 
LIMIT 0 , 30;

如果您想知道距离每个慈善机构最近的30个邮政编码和距离，那么这样的事情就可以完成工作（没有经过测试，可以解释任何错别字）。

SELECT charity, Charities.postcode, Postcodes.postcode, PostcodeDistance.distance
FROM Charities
CROSS JOIN Postcodes
INNER JOIN (SELECT PC1.postcode AS postcode1, PC2.postcode AS postcode2, (3959 * acos( cos( radians(PC1.lat) ) * cos( radians( PC2.lat ) ) * 
cos( radians( PC2.lng ) - radians(PC1.lng) ) + 
sin( radians(PC1.lat) ) * sin( radians( PC2.lat ) ) ) 
  ) AS distance 
FROM postcodes PC1
CROSS JOIN postcodes PC2) PostcodeDistance
ON Charities.postcode = PostcodeDistance.postcode1
AND Postcodes.postcode = PostcodeDistance.postcode2
HAVING distance < 30 
ORDER BY distance 
LIMIT 0 , 30;

这应该可以找到30英里范围内的慈善机构

SELECT charity, Charities.postcode, PostcodeDistance.distance
FROM Charities
INNER JOIN (
SELECT PC2.postcode AS postcode2, (3959 * acos( cos( radians(PC1.lat) ) * cos( radians( PC2.lat ) ) * 
cos( radians( PC2.lng ) - radians(PC1.lng) ) + 
sin( radians(PC1.lat) ) * sin( radians( PC2.lat ) ) ) 
) AS distance 
FROM postcodes PC1
CROSS JOIN postcodes PC2
WHERE PC1.postcode='WN8'  
) PostcodeDistance
ON Charities.postcode = PostcodeDistance.postcode2
WHERE PostcodeDistance.distance < 30 
ORDER BY PostcodeDistance.distance 
LIMIT 0 , 30;

从邮政编码表（MySql）获取经度和纬度并加入另一个表

1 个答案: