COUNT（*）和有

Question

以下查询为我提供了一列距离。 但我需要的只是匹配距离的结果数，而不是距离本身。不能使用Subselect。

SELECT 
( 6368 * SQRT(2*(1-cos(RADIANS(loc_lat)) * cos(0.899945742869) * (sin(RADIANS(`loc_lon`)) * sin(0.14286767838) + cos(RADIANS(`loc_lon`)) * cos(0.14286767838)) - sin(RADIANS(loc_lat)) * sin(0.899945742869))) ) AS Distance

FROM ...

WHERE ...

HAVING Distance > 0 AND Distance <= 25 

Answer 1

您只需将距离计算移至where子句：

SELECT COUNT(*) FROM ...

WHERE ( 6368 * SQRT(2*(1-...) BETWEEN 0 AND 25

Answer 2

如果您不需要距离，只需要计数，这可能会有效：

SELECT Count(*)
FROM ...
WHERE ... AND
  (6368 * SQRT(2*(1-cos(RADIANS(loc_lat)) * cos(0.899945742869) *
  (sin(RADIANS(`loc_lon`)) * sin(0.14286767838) + cos(RADIANS(`loc_lon`)) *
   cos(0.14286767838)) - sin(RADIANS(loc_lat)) * sin(0.899945742869)))
  ) BETWEEN 0 AND 25

Answer 3

这将给出totalResults，你可以丢弃另一列。

 SELECT COUNT(*) totalResults, 
    ( 6368 * SQRT(2*(1-cos(RADIANS(loc_lat)) * cos(0.899945742869) * (sin(RADIANS(`loc_lon`)) * sin(0.14286767838) + cos(RADIANS(`loc_lon`)) * cos(0.14286767838)) - sin(RADIANS(loc_lat)) * sin(0.899945742869))) ) AS Distance

    FROM ...

    WHERE ...

    HAVING Distance > 0 AND Distance <= 25