如何使用Facebook API从Latitude/Longitude获取城市的地方ID

Question

我需要为很多纬度/长点找到这个城市的Facebook位置.实际的点数是指个人地址,因此没有确切的地方ID可供查找,如同业务一样.

为了测试,我一直在寻找CO的Red Feather Lakes镇.

• 图搜索功能会返回很多地方,但不返回城市示例

• 原始FQL不允许您通过lat/long搜索,并且无论如何都没有"附近"的概念.例

• 按ID的FQL查询显示至少有一个"显示子文本"字段,表示该对象是一个城市.例

谢谢你的帮助.我有超过80年的父亲的日期和地理标记照片,他很想在他的时间轴上看到!

编辑

城市不在place表中,它们只在page表中.有一个未记录的distance()FQL函数,但它只适用于place表.(通过这个SO答案.)

这有效:

SELECT name,description,geometry,latitude,longitude, display_subtext 
FROM place 
WHERE distance(latitude, longitude, "40.801985", "-105.593719") < 50000

但这会给出错误"距离在表格页面中无效":

SELECT page_id,name,description,type,location 
FROM page 
WHERE distance(
  location.latitude,location.longitude,
  "40.801985", "-105.593719") < 50000

Answer 1

这是一个光荣的黑客，但这段代码是有效的。诀窍是进行两个查询。首先，我们寻找靠近我们的点的地方。这样又回来了很多营业场所。然后，我们获取这些地方之一的城市page，并使用它在表中查找该城市的页面。城市似乎有一个标准的命名约定，但美国和非美国城市的命名约定有所不同。

一些小城市在表中具有不同的拼写place，因此代码循环遍历返回的位置，直到在page表中找到匹配项。

$fb_token = 'YOUR_TOKEN';

// Red Feather Lakes, Colorado
$lat = '40.8078';
$long = '-105.579';
// Karlsruhe, Germany
$lat = '49.037868';
$long = '8.350124';

$states_arr  = array('AL'=>"Alabama",'AK'=>"Alaska",'AZ'=>"Arizona",'AR'=>"Arkansas",'CA'=>"California",'CO'=>"Colorado",'CT'=>"Connecticut",'DE'=>"Delaware",'FL'=>"Florida",'GA'=>"Georgia",'HI'=>"Hawaii",'ID'=>"Idaho",'IL'=>"Illinois", 'IN'=>"Indiana", 'IA'=>"Iowa",  'KS'=>"Kansas",'KY'=>"Kentucky",'LA'=>"Louisiana",'ME'=>"Maine",'MD'=>"Maryland", 'MA'=>"Massachusetts",'MI'=>"Michigan",'MN'=>"Minnesota",'MS'=>"Mississippi",'MO'=>"Missouri",'MT'=>"Montana",'NE'=>"Nebraska",'NV'=>"Nevada",'NH'=>"New Hampshire",'NJ'=>"New Jersey",'NM'=>"New Mexico",'NY'=>"New York",'NC'=>"North Carolina",'ND'=>"North Dakota",'OH'=>"Ohio",'OK'=>"Oklahoma", 'OR'=>"Oregon",'PA'=>"Pennsylvania",'RI'=>"Rhode Island",'SC'=>"South Carolina",'SD'=>"South Dakota",'TN'=>"Tennessee",'TX'=>"Texas",'UT'=>"Utah",'VT'=>"Vermont",'VA'=>"Virginia",'WA'=>"Washington",'DC'=>"Washington D.C.",'WV'=>"West Virginia",'WI'=>"Wisconsin",'WY'=>"Wyoming");

$place_search = json_decode(file_get_contents('https://graph.facebook.com/search?type=place&center=' . $lat . ',' . $long . '&distance=10000&access_token=' . $fb_token));
foreach($place_search->data as $result) {
  if ($result->location->city) {
    $city = $result->location->city;
    $state = $result->location->state;
    $country = $result->location->country;
    if ($country=='United States') {
      $city_name = $city . ', ' . $states_arr[$state]; // e.g. 'Chicago, Illinois'
    }
    else {
      $city_name = $city . ', ' . $country; // e.g. 'Rome, Italy'
    }
    $fql = 'SELECT name,page_id,name,description,type,location FROM page WHERE type="CITY" and name="' .$city_name. '"';
    $result = json_decode(file_get_contents('https://graph.facebook.com/fql?q=' . rawurlencode($fql) . '&access_token=' . $fb_token));
    if (count($result->data)>0) {
      // We found it!
      print_r($result);
      break;
    }
    else {
      // No luck, try the next place
      print ("Couldn't find " . $city_name . "\n");
    }
  }
}