谷歌地图 - 从经纬度获取国家、地区和城市名称 (PHP)
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Google maps - get country, region and city name from longitude and latitude (PHP)
提问by Martha Smithers
I would like to get country, region and city name from longitude and latitude over Google maps API (V3?) - with PHP. I just can't manage to get it for every county, I managed to get this code but it doesn't work on other counties because the returned result is different (the only correct information is country but even that is returned in local name not English name):
我想通过 Google 地图 API（V3？）从经纬度获取国家、地区和城市名称 - 使用 PHP。我只是无法为每个县都获得它，我设法获得了这个代码，但它在其他县不起作用，因为返回的结果是不同的（唯一正确的信息是国家，但即使是以当地名称返回的也不是英文名）：
<?
$lok1 = $_REQUEST['lok1'];
$lok2 = $_REQUEST['lok2'];
$url = 'http://maps.google.com/maps/geo?q='.$lok2.','.$lok1.'&output=json';
$data = @file_get_contents($url);
$jsondata = json_decode($data,true);
if(is_array($jsondata )&& $jsondata ['Status']['code']==200)
{
$addr = $jsondata ['Placemark'][0]['AddressDetails']['Country']['CountryName'];
$addr2 = $jsondata ['Placemark'][0]['AddressDetails']['Country']['Locality']['LocalityName'];
$addr3 = $jsondata ['Placemark'][0]['AddressDetails']['Country']['Locality']['DependentLocality']['DependentLocalityName'];
}
echo "Country: " . $addr . " | Region: " . $addr2 . " | City: " . $addr3;
?>
This works great to get correct data from my country, but not for others...
这对于从我的国家/地区获取正确数据非常有效，但不适用于其他国家...
回答by srbhbarot
You can use second api(suggested by Thomas) like this:
您可以像这样使用第二个 api（由 Thomas 建议）：
$geocode=file_get_contents('http://maps.googleapis.com/maps/api/geocode/json?latlng=48.283273,14.295041&sensor=false');
$output= json_decode($geocode);
echo $output->results[0]->formatted_address;
Try this..
尝试这个..
Edit:::::
编辑：：：：：
$geocode=file_get_contents('http://maps.googleapis.com/maps/api/geocode/json?latlng=48.283273,14.295041&sensor=false');
$output= json_decode($geocode);
for($j=0;$j<count($output->results[0]->address_components);$j++){
echo '<b>'.$output->results[0]->address_components[$j]->types[0].': </b> '.$output->results[0]->address_components[$j]->long_name.'<br/>';
}
回答by Chandrika Shah
Here is my approach to the problem
这是我解决问题的方法
$lat = "22.719569";
$lng = "75.857726";
$data = file_get_contents("http://maps.google.com/maps/api/geocode/json?latlng=$lat,$lng&sensor=false");
$data = json_decode($data);
$add_array = $data->results;
$add_array = $add_array[0];
$add_array = $add_array->address_components;
$country = "Not found";
$state = "Not found";
$city = "Not found";
foreach ($add_array as $key) {
if($key->types[0] == 'administrative_area_level_2')
{
$city = $key->long_name;
}
if($key->types[0] == 'administrative_area_level_1')
{
$state = $key->long_name;
}
if($key->types[0] == 'country')
{
$country = $key->long_name;
}
}
echo "Country : ".$country." ,State : ".$state." ,City : ".$city;
回答by Harsh Patel
<?php
$deal_lat=30.469301;
$deal_long=70.969324;
$geocode=file_get_contents('http://maps.googleapis.com/maps/api/geocode/json?latlng='.$deal_lat.','.$deal_long.'&sensor=false');
$output= json_decode($geocode);
for($j=0;$j<count($output->results[0]->address_components);$j++){
$cn=array($output->results[0]->address_components[$j]->types[0]);
if(in_array("country", $cn))
{
$country= $output->results[0]->address_components[$j]->long_name;
}
}
echo $country;
?>
回答by Thomas Antony
Have you tried using their GeoCoding API?
您是否尝试过使用他们的 GeoCoding API？
https://developers.google.com/maps/documentation/geocoding/#ReverseGeocoding
https://developers.google.com/maps/documentation/geocoding/#ReverseGeocoding
Example: http://maps.googleapis.com/maps/api/geocode/json?latlng=40.714224,-73.961452&sensor=false
示例：http: //maps.googleapis.com/maps/api/geocode/json?latlng=40.714224,-73.961452&sensor=false
Edit: I have used it in many apps targeting India and it has worked pretty well.
编辑：我在许多针对印度的应用程序中使用过它，并且效果很好。
