如何按纬度和经度查找最近的位置?

问题描述 投票:3回答:2

这是loc_coordinate表结构:

下面是代码,从数据库中获取最近的位置并显示存储在数据库本身的地名。

<?php
include("config.php");
$lat = "3.107685";
$lon = "101.7624521";

        $sql="SELECT ((ACOS(SIN($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 loc_coordinate HAVING 'distance'<='10' ORDER BY 'distance' ASC";
        $stmt =$pdo->prepare($sql);
        $stmt->execute();


        while($row = $stmt->fetch())
        {
          echo $row['place'];
        }

?>

显示的错误:

致命错误:在第8行的C:\ wamp \ www \ mysite \ by_coor.php中

PDOException:在第8行的C:\ wamp \ www \ mysite \ by_coor.php中

echo $sql显示:

SELECT((ACOS(SIN(3.107685 * PI()/ 180)* SIN(lat * PI()/ 180)+ COS(3.107685 * PI()/ 180)* COS(lat * PI()/ 180)* COS ((101.7624521-lon)* PI()/ 180))* 180 / PI())* 60 * 1.1515)AS'距离'来自loc_coordinate'距离'<='10'按'距离'ASC

我不确定为什么我会收到这个错误。这是我在SQL查询中提到的网站:http://zcentric.com/2010/03/11/calculate-distance-in-mysql-with-latitude-and-longitude/

php mysql google-maps
2个回答
15
投票

试试这个

     SELECT * , (3956 * 2 * ASIN(SQRT( POWER(SIN(( $lat - LatOnTable) *  pi()/180 / 2), 2) +COS( $lat * pi()/180) * COS(LatOnTable * pi()/180) * POWER(SIN(( $long - LongOnTable) * pi()/180 / 2), 2) ))) as distance  
from yourTable  
having  distance <= 10 
order by distance

将latOnTable替换为纬度表列名称,并将longOnTable替换为表格中的经度列名称。


0
投票

这对我有用:

SELECT restoran.id,restoran.restoran , (6371 * 2 * ASIN(SQRT( POWER(SIN(( -6.9831375276568055 - restoran.lat) *  pi()/180 / 2), 2) +COS( -6.9831375276568055 * pi()/180) * COS(restoran.lat * pi()/180) * POWER(SIN(( 110.40925562381744 - restoran.lng) * pi()/180 / 2), 2) ))) as distance  from restoran having  distance <= 10 order by distance

6371数字用于转换为km

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