Question

我正在尝试从google json获取$ street，$ city和$ country字符串。它适用于我的家庭住址： http://maps.googleapis.com/maps/api/geocode/json?latlng=52.108662,6.307370&sensor=true

$url = "http://maps.googleapis.com/maps/api/geocode/json?latlng=".$lat.",".$lng."&sensor=true";
    $data = @file_get_contents($url);
    $jsondata = json_decode($data,true);
    if(is_array($jsondata) && $jsondata['status'] == "OK")
    {
          $city = $jsondata['results']['0']['address_components']['2']['long_name'];
          $country = $jsondata['results']['0']['address_components']['5']['long_name'];
          $street = $jsondata['results']['0']['address_components']['1']['long_name'];
    }

但对于具有更多数据的不同地址，如下例所示： http://maps.googleapis.com/maps/api/geocode/json?latlng=52.154184,6.199592&sensor=true 它不起作用，因为json数组中有更多数据，它使该省成为国家。

如何选择我需要的类型（long_name）？

for street：long_name其中“types”：[“route”]
for city：long_name其中“types”：[“locality”，“political”]
for country：long_name where“types”：[“country”，“political”]

地理编码JSON的示例输出：

{
   "results" : [
      {
         "address_components" : [
            {
               "long_name" : "89",
               "short_name" : "89",
               "types" : [ "street_number" ]
            },
            {
               "long_name" : "Wieck De",
               "short_name" : "Wieck De",
               "types" : [ "establishment" ]
            },
            {
               "long_name" : "Industrieweg",
               "short_name" : "Industrieweg",
               "types" : [ "route" ]
            },
            {
               "long_name" : "Zutphen",
               "short_name" : "Zutphen",
               "types" : [ "locality", "political" ]
            },
            {
               "long_name" : "Zutphen",
               "short_name" : "Zutphen",
               "types" : [ "administrative_area_level_2", "political" ]
            },
            {
               "long_name" : "Gelderland",
               "short_name" : "GE",
               "types" : [ "administrative_area_level_1", "political" ]
            },
            {
               "long_name" : "Nederland",
               "short_name" : "NL",
               "types" : [ "country", "political" ]
            },
            {
               "long_name" : "7202 CA",
               "short_name" : "7202 CA",
               "types" : [ "postal_code" ]
            }

我想我自己修了一下，特此是我的代码：

// street
foreach ($jsondata["results"] as $result) {
    foreach ($result["address_components"] as $address) {
        if (in_array("route", $address["types"])) {
            $street = $address["long_name"];
        }
    }
}
// city
foreach ($jsondata["results"] as $result) {
    foreach ($result["address_components"] as $address) {
        if (in_array("locality", $address["types"])) {
            $city = $address["long_name"];
        }
    }
}
// country
foreach ($jsondata["results"] as $result) {
    foreach ($result["address_components"] as $address) {
        if (in_array("country", $address["types"])) {
            $country = $address["long_name"];
        }
    }
}

Answer 1

您可以将数据转换为关联数组，并像

一样使用它

 $data = array();
 foreach($jsondata['results']['0']['address_components'] as $element){
     $data[ implode(' ',$element['types']) ] = $element['long_name'];
 }
 print_r($data);

 echo 'route: ' . $data['route'] . "\n";
 echo 'country: ' . $data['country political'];

Answer 2

你的代码非常好，但是在1 foreach中使用一个开关而不是重复的foreach循环不是更好吗？以下是我解析完全相同的数组的方法：

  $location = array();

  foreach ($result['address_components'] as $component) {

    switch ($component['types']) {
      case in_array('street_number', $component['types']):
        $location['street_number'] = $component['long_name'];
        break;
      case in_array('route', $component['types']):
        $location['street'] = $component['long_name'];
        break;
      case in_array('sublocality', $component['types']):
        $location['sublocality'] = $component['long_name'];
        break;
      case in_array('locality', $component['types']):
        $location['locality'] = $component['long_name'];
        break;
      case in_array('administrative_area_level_2', $component['types']):
        $location['admin_2'] = $component['long_name'];
        break;
      case in_array('administrative_area_level_1', $component['types']):
        $location['admin_1'] = $component['long_name'];
        break;
      case in_array('postal_code', $component['types']):
        $location['postal_code'] = $component['long_name'];
        break;
      case in_array('country', $component['types']):
        $location['country'] = $component['long_name'];
        break;
    }

  }

Answer 3

如果您使用邮政编码查找地址，因为我最近使用Google MAP API生成街道，城市，国家/地区，代码为：

$search_code = urlencode($postcode);
        $url = 'http://maps.googleapis.com/maps/api/geocode/json?address=' . $search_code . '&sensor=false';
        $json = json_decode(file_get_contents($url));
        if($json->results == []){
            return '';
        }
        $lat = $json->results[0]->geometry->location->lat;
        $lng = $json->results[0]->geometry->location->lng;

        //Now build the actual lookup
        $address_url = 'http://maps.googleapis.com/maps/api/geocode/json?latlng=' . $lat . ',' . $lng . '&sensor=false';
        $address_json = json_decode(file_get_contents($address_url));

        $address_data = $address_json->results[0]->address_components;
        //return $address_data = $address_json->results[0]->formatted_address;

        $street = str_replace('Dr', 'Drive', $address_data[1]->long_name);
        $town = $address_data[2]->long_name;
        $county = $address_data[3]->long_name;

        return $street.', '. $town. ', '.$county;

Answer 4

看起来像JMESpath http://jmespath.org/

这样的集合解析器的工作

给定数组

{
  "locations": [
    {"name": "Seattle", "state": "WA"},
    {"name": "New York", "state": "NY"},
    {"name": "Bellevue", "state": "WA"},
    {"name": "Olympia", "state": "WA"}
  ]
}

JMESPath：

locations[?state == 'WA'].name | sort(@) | {WashingtonCities: join(', ', @)}

产量

{
  "WashingtonCities": "Bellevue, Olympia, Seattle"
}

您必须为您的案例重写，但您了解这种语言有多强大。您可以使用composer安装PHP的JMESPath实现

使用谷歌通过反向地理编码获取街道，城市和国家

4 个答案: