Question

我计算逐渐执行此查询的坐标之间的距离。



     /*insert into tmp11(select * from clients_lite where inwork > 0 limit(5000));*/

2



     select *, cast(ATAN(SQRT(power((COS(tmp11.lat*PI()/180)*SIN(city.lon1*PI()/180-tmp11.lon*PI()/180)),2)+power((COS(city.lat1*PI()/180)
                  *SIN(tmp11.lat*PI()/180)-SIN(city.lat1*PI()/180)*COS(tmp11.lat*PI()/180)*COS(city.lon1*PI()/180-tmp11.lon*PI()/180)),2))/
                  (SIN(city.lat1*PI()/180)*SIN(tmp11.lat*PI()/180)+COS(city.lat1*PI()/180)*COS(tmp11.lat*PI()/180)*COS(city.lon1*PI()/180-tmp11.lon*PI()/180)))*6372795 as int) as distance
    into dist
    from tmp11 
    inner join  city
    on cast(ATAN(SQRT(power((COS(tmp11.lat*PI()/180)*SIN(city.lon1*PI()/180-tmp11.lon*PI()/180)),2)+power((COS(city.lat1*PI()/180)
                  *SIN(tmp11.lat*PI()/180)-SIN(city.lat1*PI()/180)*COS(tmp11.lat*PI()/180)*COS(city.lon1*PI()/180-tmp11.lon*PI()/180)),2))/
                  (SIN(city.lat1*PI()/180)*SIN(tmp11.lat*PI()/180)+COS(city.lat1*PI()/180)*COS(tmp11.lat*PI()/180)*COS(city.lon1*PI()/180-tmp11.lon*PI()/180)))*6372795 as int) between 0 and 60000

3。

/*insert into clients_res(select clients.pid, clients.owner_id, dist.country, dist.city, dist.distance, clients.inwork  from clients left outer join dist on clients.pid = dist.pid);
 */

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如何将其转换为程序？

我怎么能用Postgresql中的存储过程呢？

0 个答案: