Question

以下代码from this post对我有用，但它会返回完整地址（街道号码，街道名称，城市，州，邮政编码，国家/地区）。有没有办法让它只返回街道号码和名称而不是完整的地址？谷歌没有多大帮助。

<?
  function getaddress($lat,$lng)
  {
     $url = 'http://maps.googleapis.com/maps/api/geocode/json?latlng='.trim($lat).','.trim($lng).'&sensor=false';
     $json = @file_get_contents($url);
     $data=json_decode($json);
     $status = $data->status;
     if($status=="OK") {
       return $data->results[0]->formatted_address;
     } else {
       return false;
     }
  }

  $lat= 26.754347; //latitude
  $lng= 81.001640; //longitude
  $address= getaddress($lat,$lng);
  if($address) {
    echo $address;
  } else {
    echo "Not found";
  }
?>

Answer 1

经过大量的搜索和反复试验后，我设法得到了我需要的工作。如果有更有效的方法来做到这一点，我会喜欢这些反馈。

<?php
    function getAddress($lat, $lon) {
        $url = "http://maps.googleapis.com/maps/api/geocode/json?latlng=".
        $lat.",".$lon."&sensor=false";
        $json = @file_get_contents($url);
        $data = json_decode($json);
        $status = $data->status;
        $address = '';
        if($status == "OK"){
            foreach($data->results[0]->address_components as $address_component) {
                if(in_array('street_number', $address_component->types)) {
                    $street_number = $address_component->long_name;
                }
                if(in_array('route', $address_component->types)) {
                     $route = $address_component->long_name;
                }
            }
        }
        return $street_number." ".$route;
    }

    echo getAddress($lat,$lon);
?>

Answer 2

<!DOCTYPE HTML>

<head>
<script src="http://code.jquery.com/jquery-2.1.4.min.js"></script>
<script type="text/javascript">
var watchID;
var geoLoc;
// Get the latitude & longitude;
function showLocation(position) {
        var latitude = position.coords.latitude;
        var longitude = position.coords.longitude;
        getAddress(latitude, longitude);
}
// Handel any errors that my come
function errorHandler(err) {
  if(err.code == 1) {
    alert("Error: Access is denied!");
  }else if( err.code == 2) {
    alert("Error: Position is unavailable!");
  }
}
// Get the location of the current location settings
function getLocation(){
    $("#output").html("Getting Location. Please Wait...");
   if(navigator.geolocation){
      // timeout at 60000 milliseconds (60 seconds)
      var options = {enableHighAccuracy:true,maximumAge:30000,timeout:27000};
      geoLoc = navigator.geolocation;
      watchID = geoLoc.watchPosition(showLocation, errorHandler, options);
   }else{
      alert("Sorry, browser does not support geolocation!");
   }
}
// Get the address
function getAddress(latitude, longitude){
    $.get("http://maps.google.com/maps/api/geocode/xml?latlng=" + latitude + "," + longitude + "&sensor=false", function(data){
        $(data).find("formatted_address").each(function(){
            var unitData = $(this).text();
            $("#output").html("Latitude : " + latitude + "<br> Longitude: " + longitude + "<br>");
            unitData = unitData.split(',');
            var street_number = unitData[0];
            var street_address = unitData[1];
            $("#output").append(street_number);
            $("#output").append(street_address);
            return false;
        });     
    });
}
</script>
</head>
<html>
  <body>
<form>
    <input type="button" onclick="getLocation();" value="Get Location"/>
  </form>
<div id="output"></div>
</body>
</html>

用于街道号码的反向地理编码的PHP功能

2 个答案: