Question

此查询工作正常，但我想在此查询中使用MIN函数，因为我只想要一个距离给定当前位置的距离最近的工作站。

select *
from
(
  SELECT fc_station_mst.station_name,fc_station_mst.station_id,fc_station_mst.lat, fc_station_mst.longitude,fc_station_mst.state,
          fc_station_param_dtl.fc_date,fc_station_param_dtl.param_value,fc_parameter_mst.param_name,fc_parameter_mst.param_unit, 

  6371*(2*atan2 (sqrt(sin(((3.14*(fc_station_mst.lat-23.07))/180)/2) * 
  sin(((3.14*(fc_station_mst.lat-23.07))/180)/2)+cos((3.14*fc_station_mst.lat)/180)*cos((3.14*23.07)/180) 
  * sin(((3.14*(fc_station_mst.longitude-72.71))/180)/2)* 
  sin(((3.14*(fc_station_mst.longitude-72.71))/180)/2)),(sqrt(1-(sin(((3.14*(fc_station_mst.lat-23.07))/180)/2)
  *sin(((3.14*(fc_station_mst.lat-23.07))/180)/2) +cos((3.14*fc_station_mst.lat)/180)* cos((3.14*23.07)/180)*
  sin(((3.14*(fc_station_mst.longitude-72.71))/180)/2)* sin(((3.14*(fc_station_mst.longitude-72.71))/180)/2)))))) as distance 

  FROM fc_station_mst, fc_station_param_dtl, fc_parameter_mst 

  where fc_station_param_dtl.param_id=fc_parameter_mst.param_id and
  fc_station_mst.station_id=fc_station_param_dtl.station_id  and  fc_station_param_dtl.param_id in(1,3,5,9)  and 
  fc_station_param_dtl.hourly_forecast='24' and fc_station_param_dtl.fc_date='22-JUL-2014'
) x

where   distance <= 15 ;

Answer 1

尝试使用和限制订单：

select *
from
(
  SELECT fc_station_mst.station_name,fc_station_mst.station_id,fc_station_mst.lat, fc_station_mst.longitude,fc_station_mst.state,
      fc_station_param_dtl.fc_date,fc_station_param_dtl.param_value,fc_parameter_mst.param_name,fc_parameter_mst.param_unit, 

6371*(2*atan2 (sqrt(sin(((3.14*(fc_station_mst.lat-23.07))/180)/2) * 
sin(((3.14*(fc_station_mst.lat-23.07))/180)/2)+cos((3.14*fc_station_mst.lat)/180)*cos((3.14*23.07)/180) 
* sin(((3.14*(fc_station_mst.longitude-72.71))/180)/2)* 
sin(((3.14*(fc_station_mst.longitude-72.71))/180)/2)),(sqrt(1-(sin(((3.14*(fc_station_mst.lat-23.07))/180)/2)
*sin(((3.14*(fc_station_mst.lat-23.07))/180)/2) +cos((3.14*fc_station_mst.lat)/180)* cos((3.14*23.07)/180)*
sin(((3.14*(fc_station_mst.longitude-72.71))/180)/2)* sin(((3.14*(fc_station_mst.longitude-72.71))/180)/2)))))) as distance 

FROM fc_station_mst, fc_station_param_dtl, fc_parameter_mst 

where fc_station_param_dtl.param_id=fc_parameter_mst.param_id and
fc_station_mst.station_id=fc_station_param_dtl.station_id  and    fc_station_param_dtl.param_id in(1,3,5,9)  and 
fc_station_param_dtl.hourly_forecast='24' and fc_station_param_dtl.fc_date='22-JUL-2014'
) x

where  x.distance <= 15 order by x.distance asc limit 1;

在查询中使用MIN函数

1 个答案: