我需要让这个人使用API在Twitter上关注另一个人。
我知道twitter已迁移到v1.1
我使用这些API来获取两个人之间的关系,
https://api.twitter.com/1.1/friendships/exists.json?screen_name_a=Person1&screen_name_b=Person2
https://api.twitter.com/1.1/friendships/lookup.json?screen_name=Person1,Person2
但我得到的最终结果是,
{
errors = (
{
code = 215;
message = "Bad Authentication data";
}
);
}
是否有其他确切的API可以找到我的解决方案。
任何人都帮我找到一个人跟随另一个人。
答案 0 :(得分:1)
尝试使用此代码,您可以获得所期望的内容,
-(void) followUsOnTwitter {
ACAccountStore *accountStore = [[ACAccountStore alloc] init];
ACAccountType *accountType = [accountStore accountTypeWithAccountTypeIdentifier:ACAccountTypeIdentifierTwitter];
[accountStore requestAccessToAccountsWithType:accountType withCompletionHandler:^(BOOL granted, NSError *error) {
if(granted) {
// Get the list of Twitter accounts.
NSArray *accountsArray = [accountStore accountsWithAccountType:accountType];
// For the sake of brevity, we'll assume there is only one Twitter account present.
// You would ideally ask the user which account they want to tweet from, if there is more than one Twitter account present.
if ([accountsArray count] > 0) {
// Grab the initial Twitter account to tweet from.
ACAccount *twitterAccount = [accountsArray objectAtIndex:0];
NSMutableDictionary *tempDict = [[NSMutableDictionary alloc] init];
[tempDict setValue:@"abcd" forKey:@"screen_name"]; // Change the Value String
[tempDict setValue:@"true" forKey:@"follow"];
TWRequest *postRequest = [[TWRequest alloc] initWithURL:[NSURL URLWithString:@"https://api.twitter.com/1.1/friendships/create.json"]
parameters:tempDict
requestMethod:TWRequestMethodPOST];
[postRequest setAccount:twitterAccount];
[postRequest performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) {
NSString *output = [NSString stringWithFormat:@"HTTP response status: %i", [urlResponse statusCode]];
NSLog(@"%@", output);
if (urlResponse.statusCode == 200)
{
// follow by usr
}
}];
}
}
}];
}
答案 1 :(得分:0)
你可以查看这个答案: 在v1.1 API中,他们已使用friendships / lookup API替换它,您可以在其中指定最多100个用户的逗号分隔列表,它将告诉您身份验证用户与指定的每个用户之间的连接< / p>