php mysql 根据经纬度计算距离和排序

 

复制代码
#1.两点距离(1.4142135623730951)
select st_distance(point(0,0),point(1,1));
select st_distance(point (120.10591, 30.30163),point(120.13026,30.25961)) as distance HAVING distance>0 ORDER BY distance;
mysql 5.6 添加

#2.两点球面距离(157249.0357231545m)
select st_distance_sphere(point(0,0),point(1,1)) as distance HAVING distance>0 ORDER BY distance;
select st_distance_sphere(point (120.10591, 30.30163),point(120.13026,30.25961)) as distance HAVING distance>0 ORDER BY distance;
This function was added in MySQL 5.7.6.
复制代码

 

第一个函数是计算平面坐标系下,两点的距离,就是

  • formula

 如果用于计算地球两点的距离,带入的参数是角度(经纬度),则计算的单位也是相差的角度,用此角度计算距离不准。纬度距离约111km每度,经度距离在赤道平面上是111km每度,随纬度的升高逐渐降低为0。

 

第二个函数是计算球面距离的公式,传入的参数是经纬度(经度-180~180,纬度-90~90),返回的值以m为单位的距离。

ST_Distance_Sphere(g1g2 [, radius])

 

 

如果mysql版本不支持上述函数怎么办?自己实现喽!下面是我自己写的球面距离函数

复制代码
delimiter //
drop function if exists Spherical_Distance;
create function Spherical_Distance(jin1 double,wei1 double,jin2 double,wei2 double) returns double
NO SQL
BEGIN
  declare j1 double;
  declare w1 double;
  declare j2 double;
  declare w2 double;
  declare R double;
  set j1 = jin1*PI()/180;
  set w1 = wei1*PI()/180;
  set j2 = jin2*PI()/180;
  set w2 = wei2*PI()/180;
  set R = 6370986;
  return R*acos(cos(w1)*cos(w2)*cos(j1-j2)+sin(w1)*sin(w2));        
END
//
delimiter ;
复制代码

调用方式

select Spherical_Distance(120.10591,30.30163,120.13026,30.25961);

计算出的值和st_distance_sphere函数计算结果相差不大。

 

另外的求经纬度距离方法:

复制代码
  $lat = trim($_POST['lat']);
        $lng = trim($_POST['lng']);
        $distance = "
        ROUND(
        6378.138 * 2 * ASIN(
            SQRT(
                POW(
                    SIN(
                        (
                            '$lat' * PI() / 180 - lat * PI() / 180
                        ) / 2
                    ),
                    2
                ) + COS(40.0497810000 * PI() / 180) * COS(lat * PI() / 180) * POW(
                    SIN(
                        (
                            '$lng' * PI() / 180 - lng * PI() / 180
                        ) / 2
                    ),
                    2
                )
            )
        ) * 1000 )  AS distance ";
        // 内容
        $sql = <<<doc
select id,title as shop_name,create_at,master_title,lng,lat,reg_address,province,city,area,address,start_at,end_at,phone,business_name,business_phone,pics,club_id,{$distance}
from jiazhen_shop_info as jsi
where 1=1 order by distance ASC,id desc
doc;    
复制代码

 

posted @   一个人的孤独自白  阅读(1166)  评论(0编辑  收藏  举报
编辑推荐:
· 基于Microsoft.Extensions.AI核心库实现RAG应用
· Linux系列:如何用heaptrack跟踪.NET程序的非托管内存泄露
· 开发者必知的日志记录最佳实践
· SQL Server 2025 AI相关能力初探
· Linux系列:如何用 C#调用 C方法造成内存泄露
阅读排行:
· 震惊!C++程序真的从main开始吗?99%的程序员都答错了
· 【硬核科普】Trae如何「偷看」你的代码?零基础破解AI编程运行原理
· 单元测试从入门到精通
· 上周热点回顾(3.3-3.9)
· winform 绘制太阳,地球,月球 运作规律
历史上的今天:
2018-12-03 sublime Text 正则表达式 替换文本内容使用介绍
点击右上角即可分享
微信分享提示