PHP从经度和纬度获取地址

Question

我创建了一个代码来获取用户的经度和纬度，这些变量存储在 mysql 数据库中。

我正在使用下面的代码来获取经度和纬度的地址。

在 stackoverflow 中搜索后，我找到了这个脚本，但它没有给我任何错误，也没有结果。

我尝试打印变量 $geolocation，我打印的结果像这样 33.8943186,35.505271199999996 但我没有完整的地址:(

任何人都可以帮我解决我的问题并获得用户的确切地址吗？？！！

 $geolocation = $latitude.','.$longitude;
     $request ='http://maps.googleapis.com/maps/api/geocode/json?latlng='.$geolocation.'';
    $file_contents = file_get_contents($request);
    $json_decode = json_decode($file_contents);
    if(isset($json_decode->results[0])) {
        $response = array();
        foreach($json_decode->results[0]->address_components as $addressComponet) {
            if(in_array('political', $addressComponet->types)) {
                    $response[] = $addressComponet->long_name;
print_r($response); // no error
            }
        }

        if(isset($response[0])){ $first  =  $response[0];  } else { $first  = 'null'; }
        if(isset($response[1])){ $second =  $response[1];  } else { $second = 'null'; } 
        if(isset($response[2])){ $third  =  $response[2];  } else { $third  = 'null'; }
        if(isset($response[3])){ $fourth =  $response[3];  } else { $fourth = 'null'; }
        if(isset($response[4])){ $fifth  =  $response[4];  } else { $fifth  = 'null'; }

        if( $first != 'null' && $second != 'null' && $third != 'null' && $fourth != 'null' && $fifth != 'null' ) {
            echo "<br/>Address:: ".$first;
            echo "<br/>City:: ".$second;
            echo "<br/>State:: ".$fourth;
            echo "<br/>Country:: ".$fifth;
print_r($response);// No error
        }
        else if ( $first != 'null' && $second != 'null' && $third != 'null' && $fourth != 'null' && $fifth == 'null'  ) {
            echo "<br/>Address:: ".$first;
            echo "<br/>City:: ".$second;
            echo "<br/>State:: ".$third;
            echo "<br/>Country:: ".$fourth;
        }
        else if ( $first != 'null' && $second != 'null' && $third != 'null' && $fourth == 'null' && $fifth == 'null' ) {
            echo "<br/>City:: ".$first;
            echo "<br/>State:: ".$second;
            echo "<br/>Country:: ".$third;
        }
        else if ( $first != 'null' && $second != 'null' && $third == 'null' && $fourth == 'null' && $fifth == 'null'  ) {
            echo "<br/>State:: ".$first;
            echo "<br/>Country:: ".$second;
        }
        else if ( $first != 'null' && $second == 'null' && $third == 'null' && $fourth == 'null' && $fifth == 'null'  ) {
            echo "<br/>Country:: ".$first;
        }
      }
print_r($response);// error

Answer 1

为了得到完整的地址，试试这个。

$latitude = 38.872292;
$longitude = -76.9698557;
$geocode=file_get_contents('https://maps.googleapis.com/maps/api/geocode/json?latlng=38.872292,-76.9698557&sensor=false&key=YOUR_KEY_HERE');
$output= json_decode($geocode);
$formattedAddress = @$output->results[0]->formatted_address;

该 URL 将在执行时返回 JSON 响应。您需要将其转换为对象/数组。在我的情况下，它是一个对象。只需遍历数组并获取所需的密钥，在本例中为formatted_address。

Answer 2

$lat=52.722452;
$lng=1.099504;

$geolocation = $lat.','.$lng;
$request ='http://maps.googleapis.com/maps/api/geocode/json?latlng='.$geolocation.'';
$json = json_decode( file_get_contents( $request ) );

$results  = $json->results[0];
$addr = $results->formatted_address;
$comp = $results->address_components;

$response=array();

foreach( $comp as $i => $obj ){
    if( in_array( 'political', $obj->types ) ) $response[]=$obj->long_name;
}

echo $addr;
print_r( $response );

Answer 3

您必须打印出 $response[]!
$geolocation 只是代码中定义的 long 和 lat。
如需更多帮助，请查看 Google 文档here