php - 根据php中的一个邮政编码找到最近的邮政编码

Question

我有一个包含所有州邮政编码的数据库。我在我的数据库中配置了它，字段是邮政编码、纬度、日志。如果我在我的表格中提供一个邮政编码并提交，那么我必须通过指定有限的公里数来获得最近的邮政编码。是否有任何 php 代码可以执行此操作？请帮助我吗？

我找到了一些代码来获取距离

<?php
 function getDistance($latitude1, $longitude1, $latitude2, $longitude2) {  
    $earth_radius = 6371;  

    $dLat = deg2rad($latitude2 - $latitude1);  
    $dLon = deg2rad($longitude2 - $longitude1);  

    $a = sin($dLat/2) * sin($dLat/2) + cos(deg2rad($latitude1)) * cos(deg2rad($latitude2)) * sin($dLon/2) * sin($dLon/2);  
    $c = 2 * asin(sqrt($a));  
    $d = $earth_radius * $c;  

    return $d;  
}  

?>

像这样如何根据一个邮政编码找到最近的邮政编码。

Answer 1

您可以使用 sql 查询来选择距离内的所有邮政编码 ##

    $distance=$_POST['specified_distance'];
        $search_postcode=$_POST['specified_postcode'];
        //use search postcode to get the row form database assume it to be searchrow
        $latitude=$searchrow['lat'];
        $longitude=$searchrow['long'];
        $sql = "SELECT *,(((acos(sin((".$latitude."*pi()/180)) * sin((`lat`*pi()/180))+cos((".$latitude."*pi()/180)) 
        * cos((`lat`*pi()/180)) * cos(((".$longitude."- `long`)*pi()/180))))*180/pi())*60*1.1515*1.609344) 
        as distance
        FROM `MyTable`
        WHERE distance <= ".$distance;

        $result = $conn->query($sql);

        if ($result->num_rows > 0) {
            $postcode_rows=array();
            while($row = $result->fetch_assoc()) {
                $postcode_rows[$row['distance']]=$row; //get each row to associative //array with distance as key
            }
 //do array krsort to get the least distance row;
        $sorted_postcodes_rows=krsort($postcodes);
        $selected_postcode_row=array_shift($sorted_postcodes); //get the nearest post //code
        echo =$nearest_postcode=$selected_postcode_row['postcode'];
        } else {
            echo "No nearest postalcode";
        }