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我有一个带有字段的慈善表：charity、postcode 和一个带有字段的邮政编码表：postcode、lat、lng

我想从网页发布邮政编码并找到最近的慈善机构

我是一个 mysql 初学者，所以我有点迷茫，但我一直在尝试使用连接和子查询的各种想法，但都不起作用（我要么遇到语法错误，要么'操作数应该包含 1 列'下面的代码）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;

我在这里看到了很多例子，其中 lat2 和 lng2 是从发布的值而不是从数据库中的表中获得的。

示例中的ps'where postcode='WN8'仅用于测试

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1 回答 1

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不确定上述 SQL 遇到了什么错误。

但是尝试这个小调整，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;

如果您想知道最近的 30 个邮政编码以及与每个慈善机构的距离，那么这样的事情就可以完成（未经测试，请原谅任何拼写错误）。

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;

这应该会为您找到 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;
于 2012-11-21T10:29:34.003 回答