Facebook开放图表帖子的网址是什么?

时间:2011-01-18 21:53:15

标签: facebook facebook-graph-api

给定图搜索返回的帖子ID,例如:186173001411937

是否有一个网址链接到Facebook的帖子? 以下网址不起作用: http://www.facebook.com/post.php?id=186173001411937

5 个答案:

答案 0 :(得分:48)

我发现,对于图表ID 1099696306_140549259338782,链接是这样构建的: http://www.facebook.com/1099696306/posts/140549259338782

答案 1 :(得分:21)

老实说,我发现这样做的最简单方法就是:

"http://www.facebook.com/" + postId

postId只是帖子的直接ID(186173001411937),而不是userid_postid变体。

答案 2 :(得分:5)

使用图形api v2.5,您可以使用posts对象的permalink_url字段。

即:

www.facebook.com/v2.5/{pagename}/?fields=posts{permalink_url,message,story,created_time,id}

将返回

"posts": {
   "data": [
   {
    "permalink_url": "https://www.facebook.com/etsmtl/posts/10153868925494376",
    "message": "Le Club Cedille organise le prochain Linux-Meetup ce soir à l'ÉTS. Au programme : conférence de James Shubin, ingénieur logiciel sénior chez Red Hat.",
    "created_time": "2016-03-01T15:23:11+0000",
    "id": "8632204375_10153868925494376"
   }, ... }

答案 3 :(得分:2)

对于面向公众的页面帖子,请从Facebook Graph API返回 ID ,例如 12345678_12345678 并将其附加到 facebook.com ,例如https://www.facebook.com/12345678_12345678。当您访问该页面时,帖子也会突出显示

答案 4 :(得分:0)

 public <T>getPagePosts(string pageId, string access_token, int limit)
        {
            var client = new RestClient("https://graph.facebook.com");
            var request = new RestRequest(Method.GET);
            var fields = "posts{permalink_url,picture,message,story,created_time,id}";
            request.Resource = "{version}/{object_id}/";
            request.RequestFormat = DataFormat.Json;
            request.JsonSerializer.ContentType = "application/json;";
            request.AddParameter("access_token", access_token);
            request.AddParameter("version", "v2.10", ParameterType.UrlSegment);
            request.AddParameter("object_id", pageId, ParameterType.UrlSegment);
            request.AddParameter("limit", limit);
            request.AddParameter("fields", fields);
            var response = client.Execute(request);
            var result = JsonConvert.DeserializeObject<T>(response.Content);
            return result;
        }