我正在使用STTwitter演示应用程序在授权后从演示应用程序向我的Twitter帐户发布推文和图像。我正在尝试下面:
- (void)setOAuthToken:(NSString *)token oauthVerifier:(NSString *)verifier {
[_twitter postAccessTokenRequestWithPIN:verifier successBlock:^(NSString *oauthToken, NSString *oauthTokenSecret, NSString *userID, NSString *screenName) {
_loginStatusLabel.text = [NSString stringWithFormat:@"%@ (%@) %@," , screenName];
STTwitterAPI *twitter = [STTwitterAPI twitterAPIWithOAuthConsumerKey:_consumerKeyTextField.text consumerSecret:_consumerSecretTextField.text oauthToken:_twitter.oauthAccessToken oauthTokenSecret:_twitter.oauthAccessTokenSecret];
[twitter verifyCredentialsWithUserSuccessBlock:^(NSString *username, NSString *userID){
[self.twitter postStatusesUpdate:@"test" inReplyToStatusID:nil latitude:nil longitude:nil placeID:nil displayCoordinates:nil trimUser:nil autoPopulateReplyMetadata:nil excludeReplyUserIDsStrings:nil attachmentURLString:nil useExtendedTweetMode:nil successBlock: ^(NSDictionary *status) {
NSLog(@"twitter post success");
}
errorBlock:^(NSError *error) {
NSLog(@"twitter post failed %@",error.localizedDescription);
}];
}
} errorBlock:^(NSError *error) {
_loginStatusLabel.text = [error localizedDescription];
NSLog(@"-- %@", [error localizedDescription]);
}];
}
使用postStatusesUpdate方法的简单推文也不会发布到我的推特账号。它总是输入失败块(推文发布失败)。上面的方法是从appdelegate调用的。
编辑: 我使用SLRequest成功在Twitter上发布内容,但如果没有登录在Twitter本机应用程序中,则无法使用我的应用再次登录到Twitter。以上代码适用于Twitter的授权但不发布内容,SLRequest适用于发布内容但不授权。任何想法?
答案 0 :(得分:0)
我得到了解决方案,所以想在这里发帖作为答案:
我使用TWTRAPIClient并使用userid.And初始化它,名为TWTRAPIClient sendTwitterRequest
NSString *statusesShowEndpoint = @"https://upload.twitter.com/1.1/media/upload.json";
NSDictionary *params = @{@"media" : imageString};
NSError *clientError;
NSURLRequest *request = [client URLRequestWithMethod:@"POST" URL:statusesShowEndpoint parameters:params error:&clientError];
if (request) {
[client sendTwitterRequest:request completion:^(NSURLResponse *response, NSData *data, NSError *connectionError) {
if (data) {
NSLog(@"response %@",response);
NSError *jsonError;
NSDictionary *json = [NSJSONSerialization JSONObjectWithData:data options:0 error:&jsonError];
NSLog(@"Media ID : %@",[json objectForKey:@"media_id_string"]);
NSString *mediaID =[json objectForKey:@"media_id_string"];
if (mediaID!=nil) {
NSString *statusesShowEndpoint = @"https://api.twitter.com/1.1/statuses/update.json";
NSDictionary *message = @{@"status": websiteURL, @"wrap_links": @"true", @"media_ids": mediaID};
NSError *clientError;
NSURLRequest *request = [client URLRequestWithMethod:@"POST" URL:statusesShowEndpoint parameters:message error:&clientError];
[client sendTwitterRequest:request completion:^(NSURLResponse *response, NSData *data, NSError *connectionError){
if (data) {
NSLog(@"response %@",response);
NSError *jsonError;
NSDictionary *json = [NSJSONSerialization JSONObjectWithData:data options:0 error:&jsonError];
}
else {
NSLog(@"Error: %@", connectionError);
}
}];
}
}
else {
NSLog(@"Error: %@", connectionError);
}
}];
}
else {
NSLog(@"Error: %@", clientError);
}