iOS SDK 中的 Facebook FQL 多重查询

Question

我正在寻找一些有关如何通过 iOS SDK 向 Facebook 发送多查询 fql 请求的文档或示例代码。 我找到了一些较旧的样本，但它们对我不起作用。 我已经用查询填充了 NSDictionary ，并尝试通过以下方式发送请求 [[FBRequest requestWithDelegate:self] call:@"facebook.fql.multiquery" params:params];，如我读过的示例中所述，但我收到“无法识别的选择器发送到类”错误，我也无法在 FBRequest.h 中找到“requestWithDelegate”方法。它已被弃用吗？ 对此的一些帮助将非常感激！

Answer 1

// Construct your FQL Query 
NSString* fql = [NSString stringWithFormat:@"SELECT uid, name, hometown_location from user where uid IN (SELECT uid2 FROM friend WHERE uid1=%lld) and hometown_location != '' ORDER BY rand() limit 4", facebookId];

// Create a params dictionary
NSMutableDictionary* params = [NSMutableDictionary dictionaryWithObject:fql forKey:@"query"];

// Make the request 
[facebook requestWithMethodName:@"fql.query" andParams:params andHttpMethod:@"GET" andDelegate:self];

委托 FBRequestDelegate 将随响应一起调用。根据您的需要，您将执行以下操作：

- (void)request:(FBRequest*)request didLoad:(id)result {
    NSLog(@"didLoad() received result: %@", result);
}

Facebook 对象的创建方式如下：

Facebook * facebook = [[Facebook alloc] initWithAppId:kAppID];

有关如何设置的更多信息可以在此处找到：
https://developers.facebook.com/docs/guides/mobile/

并且取决于对于 FQL 查询，您需要获得用户的权限。请参阅此处的权限列表：
https://developers.facebook.com/docs/authentication/permissions/

有这里还有一个 FQL 查询的测试控制台：
https://developers.facebook.com/docs/reference/rest/fql。 query/

如果你想执行多重查询而不是单个查询，它看起来像：

NSString* fql1 = [NSString stringWithFormat:
    @"select uid2 from friend where uid1 == %lld order by rand() limit 10", facebookId];
NSString* fql2 = [NSString stringWithFormat:
    @"select uid, name from user where uid in (select uid2 from #queryID)"];
NSString* fql3 = [NSString stringWithFormat:
    @"select uid, status_id, message from status where uid in (select uid from #queryName) limit 1"];
NSString* fql = [NSString stringWithFormat:
    @"{\"queryID\":\"%@\",\"queryName\":\"%@\",\"queryStatus\":\"%@\"}",fql1,fql2,fql3];

NSDictionary* params = [NSDictionary dictionaryWithObject:fql forKey:@"queries"];   

[facebook call:@"fql.multiquery" params:params];