Php SQL查询MySQL

我在下面执行SQL查询时遇到以下错误。你认为哪里不对

#1054 - Unknown column 'distance' in 'where clause'
SELECT longitude, latitude, firstname, surname, profile_pic, facebook_id, 6371 * ACos( Cos( RADIANS( users.latitude ) ) * Cos( RADIANS( 23 ) ) * Cos( RADIANS( 23 ) - RADIANS( users.longitude ) ) + Sin( RADIANS( users.latitude ) ) * Sin( RADIANS( 23 ) ) ) AS distance
FROM users
WHERE distance >=1000
ORDER BY distance
LIMIT 20 

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where子句在别名之前求值

SELECT longitude, latitude, firstname, surname, profile_pic, facebook_id, distance 
FROM 
  (SELECT longitude, latitude, firstname, surname, profile_pic, facebook_id, 6371 * ACos( Cos( RADIANS( users.latitude ) ) * Cos( RADIANS( 23 ) ) * Cos( RADIANS( 23 ) - RADIANS( users.longitude ) ) + Sin( RADIANS( users.latitude ) ) * Sin( RADIANS( 23 ) ) ) AS distance
   FROM users
  ) as u1
  WHERE distance >=1000
  ORDER BY distance
  LIMIT 20

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使用HAVING子句代替WHERE子句。WHERE子句中不存在别名distance

SELECT longitude, latitude, firstname, surname, profile_pic, facebook_id, 6371 * ACos( Cos( RADIANS( users.latitude ) ) * Cos( RADIANS( 23 ) ) * Cos( RADIANS( 23 ) - RADIANS( users.longitude ) ) + Sin( RADIANS( users.latitude ) ) * Sin( RADIANS( 23 ) ) ) AS distance
FROM users
HAVING distance >=1000
ORDER BY distance
LIMIT 20 

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您可以使用内部查询，如

SELECT * 
FROM 
  (
   SELECT longitude, latitude, firstname, surname, profile_pic, facebook_id, 6371 * ACos( Cos( RADIANS( users.latitude ) ) * Cos( RADIANS( 23 ) ) * Cos( RADIANS( 23 ) - RADIANS( users.longitude ) ) + Sin( RADIANS( users.latitude ) ) * Sin( RADIANS( 23 ) ) ) AS distance
   FROM Users
  ) as u1
WHERE distance >=1000
ORDER BY distance
LIMIT 20

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距离不是一个列名称这就是为什么我喜欢在LINQ中强制使用的格式，而不是从WHERE SELECT中。我不知道WHERE子句首先计算+1谢谢，但是我现在在您的查询中发现了这个错误#1248-每个派生表都必须有自己的alias@Jae将别名添加到子查询中。