Question

这是我的SQLFiddle，这是我的程序

我想将此过程重写为简单操作。

它应该将输入作为Lat，Lng并且相对于给定的Lat和Long返回最近的driverid，lat，long（由desc排序并取第一个记录）。它包含一些我不需要的表。

（问题是我无法找到哪个字段是必需的，哪个不是）

DELIMITER //
CREATE  PROCEDURE `driver_latlong`(IN `lat` DECIMAL(20,9), IN `lng` DECIMAL(20,9))
BEGIN
SET @sql = CONCAT('SELECT sd.DriverId, sd.Lat, sd.Long,  sd1.RadiusOfService, @dist:= ( 6371 * acos( cos( radians(',lat,') ) * cos( radians( sd.Lat ) ) * cos( radians( sd.Long ) - radians(',lng,') ) + sin( radians(',lat,') ) * sin( radians( sd.Lat ) ) ) ) AS distance 
        , Case When @dist <= 1 Then sd1.DeliveryFeeOne When @dist<= 2 Then sd1.DeliveryFeeTwo When @dist<= 3 Then sd1.DeliveryFeeThree Else sd1.DeliveryFeeMore End as deliveryfee
        , Case When @dist <= 1 Then sd1.DeliveryTimeOne When @dist<= 2 Then sd1.DeliveryTimeTwo When @dist<= 3 Then sd1.DeliveryTimeThree Else sd1.DeliveryTimeMore End as deliverytime
        , sd.ShopName, sd.Address, sd.LogoFile FROM driver_latlong sd LEFT JOIN  WHERE sd.Status = \'1\' and sd.Lat != 0 and sd.Long != 0 AND ( 6371 * acos( cos( radians(',lat,') ) * cos( radians( sd.Lat ) ) * cos( radians( sd.Long ) - radians(',lng,') ) + sin( radians(',lat,') ) * sin( radians( sd.Lat ) ) ) ) <= sd1.RadiusOfService  ORDER BY distance ASC;');                
                PREPARE stmt FROM @sql;
                EXECUTE stmt;
                DEALLOCATE PREPARE stmt;
    END //
    DELIMITER ;

注意：实际上我从之前的项目中得到了这个程序，所以不需要其他表只我们引用的表是driver_latlong

Answer 1

使用以下查询获取结果并使用您的硬编码值代替$ lat和$ lon

 SELECT ((ACOS(SIN($lat * PI() / 180) * SIN(lat * PI() / 180) + COS($lat * PI() / 180) * COS(lat * PI() / 180) * COS(($lon – lon) * PI() / 180)) * 180 / PI()) * 60 * 1.1515) AS `distance` FROM `members` HAVING `distance`<5

查找所提供纬度和经度的最近距离记录的过程

1 个答案: