php 如何获取任何地址的经度和纬度？
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How to get longitude and latitude of any address?
提问by Ajay_kumar
Hello what is the formulla to get longitude and latitude of any address , if i have street_name,State_name,City_name,Country_name and zip code in PHP? Thanks
您好，如果我在 PHP 中有 street_name、State_name、City_name、Country_name 和邮政编码，那么获取任何地址的经度和纬度的公式是什么？谢谢
回答by Surya prakash Patel
Use the following code for getting lat and long using php. Here are two methods:
使用以下代码使用 php 获取经纬度。这里有两种方法：
Type-1:
类型 1：
<?php
// Get lat and long by address
$address = $dlocation; // Google HQ
$prepAddr = str_replace(' ','+',$address);
$geocode=file_get_contents('https://maps.google.com/maps/api/geocode/json?address='.$prepAddr.'&sensor=false');
$output= json_decode($geocode);
$latitude = $output->results[0]->geometry->location->lat;
$longitude = $output->results[0]->geometry->location->lng;
?>
edit - Google Maps requests must be over https
编辑 - 谷歌地图请求必须通过 https
Type-2:
类型 2：
<script src="https://maps.googleapis.com/maps/api/js?v=3.exp&sensor=false"></script>
<script>
var geocoder;
var map;
function initialize() {
geocoder = new google.maps.Geocoder();
var latlng = new google.maps.LatLng(50.804400, -1.147250);
var mapOptions = {
zoom: 6,
center: latlng
}
map = new google.maps.Map(document.getElementById('map-canvas12'), mapOptions);
}
function codeAddress(address,tutorname,url,distance,prise,postcode) {
var address = address;
geocoder.geocode( { 'address': address}, function(results, status) {
if (status == google.maps.GeocoderStatus.OK) {
map.setCenter(results[0].geometry.location);
var marker = new google.maps.Marker({
map: map,
position: results[0].geometry.location
});
var infowindow = new google.maps.InfoWindow({
content: 'Tutor Name: '+tutorname+'<br>Price Guide: '+prise+'<br>Distance: '+distance+' Miles from you('+postcode+')<br> <a href="'+url+'" target="blank">View Tutor profile</a> '
});
infowindow.open(map,marker);
} /*else {
alert('Geocode was not successful for the following reason: ' + status);
}*/
});
}
google.maps.event.addDomListener(window, 'load', initialize);
window.onload = function(){
initialize();
// your code here
<?php foreach($addr as $add) {
?>
codeAddress('<?php echo $add['address']; ?>','<?php echo $add['tutorname']; ?>','<?php echo $add['url']; ?>','<?php echo $add['distance']; ?>','<?php echo $add['prise']; ?>','<?php echo substr( $postcode1,0,4); ?>');
<?php } ?>
};
</script>
<div id="map-canvas12"></div>
回答by Nana Partykar
<?php
$address = 'BTM 2nd Stage, Bengaluru, Karnataka 560076'; // Address
$apiKey = 'api-key'; // Google maps now requires an API key.
// Get JSON results from this request
$geo = file_get_contents('https://maps.googleapis.com/maps/api/geocode/json?address='.urlencode($address).'&sensor=false&key='.$apiKey);
$geo = json_decode($geo, true); // Convert the JSON to an array
if (isset($geo['status']) && ($geo['status'] == 'OK')) {
$latitude = $geo['results'][0]['geometry']['location']['lat']; // Latitude
$longitude = $geo['results'][0]['geometry']['location']['lng']; // Longitude
}
?>
回答by elsni
You need to access a geocoding service (i.e. from Google), there is no simple formula to transfer addresses to geo coordinates.
您需要访问地理编码服务（即来自 Google），没有简单的公式可以将地址转换为地理坐标。
回答by Thomas Langston
You can use the Google Maps API for that. See the blog post below for more information.
为此，您可以使用 Google Maps API。有关更多信息，请参阅下面的博客文章。
http://stuff.nekhbet.ro/2008/12/12/how-to-get-coordinates-for-a-given-address-using-php.html
http://stuff.nekhbet.ro/2008/12/12/how-to-get-coordinates-for-a-given-address-using-php.html
回答by Kyle Coots
PHP has some nice built in functions for getting geographic location. Maybe have a look here: http://php.net/manual/en/ref.geoip.php
PHP 有一些很好的内置函数来获取地理位置。也许看看这里：http: //php.net/manual/en/ref.geoip.php
According to php manual, "This extension requires the GeoIP C library version 1.4.0 or higher to be installed. You can grab the latest version from ? http://www.maxmind.com/app/cand compile it yourself."
根据 php 手册，“此扩展需要安装 GeoIP C 库版本 1.4.0 或更高版本。您可以从 ? http://www.maxmind.com/app/c获取最新版本并自行编译。”
回答by Ian
I came up with the following which takes account of rubbish passed in and file_get_contents failing....
我想出了以下内容，其中考虑了传入的垃圾和 file_get_contents 失败....
function get_lonlat( $addr ) {
try {
$coordinates = @file_get_contents('http://maps.googleapis.com/maps/api/geocode/json?address=' . urlencode($addr) . '&sensor=true');
$e=json_decode($coordinates);
// call to google api failed so has ZERO_RESULTS -- i.e. rubbish address...
if ( isset($e->status)) { if ( $e->status == 'ZERO_RESULTS' ) {echo '1:'; $err_res=true; } else {echo '2:'; $err_res=false; } } else { echo '3:'; $err_res=false; }
// $coordinates is false if file_get_contents has failed so create a blank array with Longitude/Latitude.
if ( $coordinates == false || $err_res == true ) {
$a = array( 'lat'=>0,'lng'=>0);
$coordinates = new stdClass();
foreach ( $a as $key => $value)
{
$coordinates->$key = $value;
}
} else {
// call to google ok so just return longitude/latitude.
$coordinates = $e;
$coordinates = $coordinates->results[0]->geometry->location;
}
return $coordinates;
}
catch (Exception $e) {
}
then to get the cords: where $pc is the postcode or address.... $address = get_lonlat( $pc ); $l1 = $address->lat; $l2 = $address->lng;
然后获取电源线：其中 $pc 是邮政编码或地址.... $address = get_lonlat( $pc ); $l1 = $address->lat; $l2 = $address->lng;
回答by swestrup
There is no forumula, as street names and cities are essentially handed out randomly. The address needs to be looked up in a database. Alternatively, you can look up a zip code in a database for the region that the zip code is for.
没有论坛，因为街道名称和城市基本上是随机分发的。地址需要在数据库中查找。或者，您可以在数据库中查找邮政编码所在区域的邮政编码。
You didn't mention a country, so I'm going to assume you just want addresses in the USA. There are numerous databases you can use, some free, some not.
你没有提到一个国家，所以我假设你只想要美国的地址。您可以使用许多数据库，有些是免费的，有些则不是。
You can also use the Google Maps API to have them look up an address in their database for you. That is probably the easiest solution, but requires your application to have a working internet connection at all times.
您还可以使用 Google Maps API 让他们在他们的数据库中为您查找地址。这可能是最简单的解决方案，但要求您的应用程序始终具有可用的 Internet 连接。
