【微信开发】微信小程序通过经纬度计算两地距离php代码实现

/**
 *  @desc 根据两点间的经纬度计算距离
 *  @param float $lat 纬度值
 *  @param float $lng 经度值
 *  @param $status true KM，M显示;false 只返回M
 */
function system_getdistance($lat1,$lng1,$lat2,$lng2,$status = true,$single = true)
{
    $earthRadius = 6367000;
    //approximate radius of earth in meters
    $lat1 = (floatval($lat1) * pi()) / 180;
    $lng1 = (floatval($lng1) * pi()) / 180;
    $lat2 = (floatval($lat2) * pi()) / 180;
    $lng2 = (floatval($lng2) * pi()) / 180;
    $calcLongitude = $lng2 - $lng1;
    $calcLatitude = $lat2 - $lat1;
    $stepOne = pow(sin($calcLatitude / 2),2) + cos($lat1) * cos($lat2) * pow(sin($calcLongitude / 2),2);
    $stepTwo = 2 * asin(min(1,sqrt($stepOne)));
    $calculatedDistance = $earthRadius * $stepTwo;
    if (!$single) {
        return round($calculatedDistance);
        exit;
    }
    //把大于1000米的转换成km
    if ($status) {
        $m = round($calculatedDistance) / 1000;
        return $m > 1 ? round($m,1) . "km" : ($m * 1000) . "m";
        //return round($m,2);       //这个是我在程序中用的，以为排序比较方便，大家根据需求选用
    } else {
        return round($calculatedDistance) . "m";
    }
}

    /**
     * @param Request $request
     * @return array
     * 获取分类商店，并按距离排序
     */
    function getShop(Request $request){
        $c_id = $request->get(‘c_id‘);
        $lat1 = $request->get(‘lat‘);
        $lng1 = $request->get(‘lng‘);
        $lat1U = floatval($lat1)+0.01;
        $lat1D = floatval($lat1)-0.01;
        $lng1U = floatval($lng1)+0.01;
        $lng1D = floatval($lng1)-0.01;
        $data = db(‘store‘)
            ->where(‘c_id‘,$c_id)
            ->where(‘lat‘,‘<‘,$lat1U)      //获取周围经纬度差0.01的商家距离
            ->where(‘lat‘,‘>‘,$lat1D)
            ->where(‘lng‘,$lng1U)
            ->where(‘lng‘,$lng1D)
            ->find();
        $res = json_encode($data);
        $arr1 = json_decode($res,true);
        $arr = [];
       foreach ($arr1 as $k=>$v){
            $res = $this->system_getdistance($lat1,$v[‘lat‘],$v[‘lng‘]);
            $arr1[$k][‘km‘] = $res.‘km‘;
            $arr[$k] = $res;
        }
        asort($arr);
        $arr2 = [];
        foreach($arr as $k=>$v){
            $arr2[] = $arr1[$k];
        }
        return $arr2;
    }

SELECT
    id,
    lat,
    lng,ROUND(
        6378.138 * 2 * ASIN(
            SQRT(
                POW(
                    SIN(
                        (
                            $lat * PI() / 180 - lat * PI() / 180
                        ) / 2
                    ),2
                ) + COS($lat * PI() / 180) * COS(lat * PI() / 180) * POW(
                    SIN(
                        (
                            $lng * PI() / 180 - lng * PI() / 180
                        ) / 2
                    ),2
                )
            )
        ) * 1000
    ) AS distance
FROM
    store
ORDER BY
    distance asc

select * from (
　　　　　　

SELECT
    id, lat, lng,ROUND( 6378.138 * 2 * ASIN( SQRT( POW( SIN( ( $lat * PI() / 180 - lat * PI() / 180 ) / 2 ),2 ) + COS($lat * PI() / 180) * COS(lat * PI() / 180) * POW( SIN( ( $lng * PI() / 180 - lng * PI() / 180 ) / 2 ),2 ) ) ) * 1000 ) AS distance FROM store ORDER BY distance asc

 ) as a where a.distance<=5000