Php 从json结果中提取一位数据

我试图在加载CURL时从JSON数组中提取一位数据。如果我回显结果的话，我可以告诉你这一切都很好，但是当我要去邮递镇的时候，我总是遇到一片空白？？？？我的PHP代码：

  $url = "https://maps.googleapis.com/maps/api/geocode/json?latlng=52.406822,-1.519693&sensor=true";
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    curl_setopt($ch, CURLOPT_PROXYPORT, 3128);
    curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 0);
    curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 0);
    $response = curl_exec($ch);
    curl_close($ch);
    $response_a = json_decode($response, true);
    $town = $response_a['results'][0]['address_components'][0]['types']['postal_town'];

    echo $town;

JSON结果的一个示例是：

{
   "results" : [
      {
         "address_components" : [
            {
               "long_name" : "Ringway Rudge",
               "short_name" : "A4053",
               "types" : [ "route" ]
            },
            {
               "long_name" : "Coventry",
               "short_name" : "Coventry",
               "types" : [ "locality", "political" ]
            },
            {
               "long_name" : "Coventry",
               "short_name" : "Coventry",
               "types" : [ "postal_town" ]
            },
            {
               "long_name" : "West Midlands",
               "short_name" : "West Midlands",
               "types" : [ "administrative_area_level_2", "political" ]
            },
            {
               "long_name" : "England",
               "short_name" : "England",
               "types" : [ "administrative_area_level_1", "political" ]
            },

---

这就成功了，照德文说的做了，谢谢

   $result = @file_get_contents("http://maps.googleapis.com/maps/api/geocode/json?latlng=53.406822,-1.519693&sensor=true" );
    if ($result === FALSE) {
        //manage exception from file_get_contents call
    } else {
        $geocodedinfo = json_decode($result);
        if ($geocodedinfo->status == "OK") {
            $town = "";
            foreach ($geocodedinfo->results[0]->address_components as $addrcomps) {
                if ( $addrcomps->types[0] == 'postal_town')
                    $town = $addrcomps->long_name;           
            }
        }
    }


    echo $town;

---

我在你的JSON中没有看到postal_town键…“types”：[“postal_town”]这是第三个，但这不是键，这是一个值。哦？我只想将的值设置为$city。对不起，也许我用错了，我不认为这是你怎么用的，这是你试图访问它的方式。你试图用一把钥匙进入postal_town，不知怎的，你希望它能给你带来回报？但postal_town不是键，而是类型数组中的值。