【问题标题】:Execute simple FQL query with spring-social-facebook in java在 java 中使用 spring-social-facebook 执行简单的 FQL 查询
【发布时间】:2014-01-07 09:46:54
【问题描述】:
对于 spring-social-facebook 库的实现,我将不胜感激。我正在尝试执行 FQL 查询,但 facebook 向我发送了空结果。如果我在 Facebook Graph API Explorer 中执行相同的查询,它就像一个魅力。我检查了所有权限。我的代码是这样的:
private static final String FQL_GET_FRIENDS = "SELECT uid, name FROM user WHERE uid IN(SELECT uid2 FROM friend WHERE uid1 = me())";
Facebook facebook = new FacebookTemplate(token);
List<Reference> references = facebook.userOperations().search(FQL_GET_FRIENDS);
感谢您的建议
【问题讨论】:
标签:
java
facebook
spring-social
spring-social-facebook
【解决方案1】:
好的。这对我有用:
MultiValueMap<String, String> parameters = new LinkedMultiValueMap<String, String>();
parameters.set("q", FQL_GET_FRIENDS);
Map<String, Object> resultSet = (Map<String, Object>) facebookTemplate
.fetchObject("fql", Map.class, parameters);
【解决方案2】:
我已经通过一个简单的 HTTP Get 请求对“facebookURL”进行了同样的操作,该请求带有一个具有足够权限的有效令牌,你可以使用这个 ::
private static String STR_FACEBOOK_FQL_URL = "https://api.facebook.com/method/fql.query?query=";
private static String STR_FACEBOOK_RETURN_FORMAT = "&format=json";
private static String STR_FACEBOOK_ENCODE_FORMAT = "UTF-8";
private static String STR_FACEBOOK_ACCESS_TOKEN = "&access_token=";
private static final String FQL_GET_FRIENDS = "SELECT uid, name FROM user WHERE uid IN(SELECT uid2 FROM friend WHERE uid1 = me())";
String encodedQuery = URLEncoder.encode(FQL_GET_FRIENDS, STR_FACEBOOK_ENCODE_FORMAT);
String faceBookURL = STR_FACEBOOK_FQL_URL + encodedQuery + STR_FACEBOOK_RETURN_FORMAT
+ STR_FACEBOOK_ACCESS_TOKEN + token;