【发布时间】:2013-10-05 13:36:51
【问题描述】:
我有上传照片的应用程序,比如 Instagram 应用程序。首先我的应用上传照片并返回链接。然后在状态消息中发送链接。当我在我的 iPhone 上进行测试时,一切正常,但是当我在另一部 iPhone 上进行测试时,twitter 返回代码 215 错误的身份验证数据。有什么想法可能是错的吗?谢谢回复。
- (void)tweet:(NSString *)identifier withCompletionHandler:(void (^)())completionHandler {
ACAccountStore *accountStore = [[ACAccountStore alloc] init];
// Create an account type that ensures Twitter accounts are retrieved.
ACAccountType *accountType = [accountStore accountTypeWithAccountTypeIdentifier:ACAccountTypeIdentifierTwitter];
// Request access from the user to use their Twitter accounts.
[accountStore requestAccessToAccountsWithType:accountType withCompletionHandler:^(BOOL granted, NSError *error) {
if(granted) {
if (identifier) {
// Get the list of Twitter accounts.
NSArray *accountsArray = [accountStore accountsWithAccountType:accountType];
if ([accountsArray count] > 0) {
// Grab the initial Twitter account to tweet from.
NSURL *url = [NSURL URLWithString:@"https://api.twitter.com/1.1/statuses/update.json"];
NSDictionary *dict = [NSDictionary dictionaryWithObject:[NSString stringWithFormat:@"Check out this great flick on BLA BLA %@", [BLA_URL stringByAppendingFormat:@"/%@/%@", @"flick", identifier]]
forKey:@"status"];
SLRequest *request = [SLRequest requestForServiceType:SLServiceTypeTwitter
requestMethod:SLRequestMethodPOST
URL:url
parameters:dict];
[request setAccount:[accountsArray lastObject]];
[request performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) {
if ([urlResponse statusCode] == 200) {
NSLog(@"TWEEET!");
}
else {
NSError *jsonError;
id data = [NSJSONSerialization JSONObjectWithData:responseData
options:NSJSONReadingMutableLeaves
error:&jsonError];
dispatch_async(dispatch_get_main_queue(), ^{
UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"Twitter response"
message:[NSString stringWithFormat:@"%d S: %d %@", [error code], [urlResponse statusCode], data]
delegate:self
cancelButtonTitle:@"Ok"
otherButtonTitles:nil];
NSLog(@"NOOO TWEEET!");
[alertView show];
completionHandler();
});
}
}];
...
及预检功能:
- (IBAction)twitterButtonTapped:(id)sender {
if ([self.twitterButton isSelected]) {
[self.twitterButton setSelected:NO];
}
else {
[self.twitterButton setSelected:YES];
ACAccountStore *accountStore = [[ACAccountStore alloc] init];
ACAccountType *accountType = [accountStore accountTypeWithAccountTypeIdentifier:ACAccountTypeIdentifierTwitter];
[accountStore requestAccessToAccountsWithType:accountType withCompletionHandler:^(BOOL granted, NSError *error) {
if (!granted) {
[self.twitterButton setSelected:NO];
UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"Twitter access"
message:@"Please make sure you have allowed twitter for BLA BLA in your Settings."
delegate:self
cancelButtonTitle:@"Ok"
otherButtonTitles:nil];
[alertView performSelectorOnMainThread:@selector(show) withObject:nil waitUntilDone:NO];
}
}];
}
}
【问题讨论】:
-
我遇到了同样的问题。类似的代码一直有效,直到我不得不切换到 API v 1.1
-
其他 iPhone 是否在设置中设置了 Twitter 帐户?对话框是否显示在要求您授予访问权限的位置?
-
是的,其他 iPhone 有大约 10 个 Twitter 帐户。是的,当无法访问时会出现对话。当有人点击“在 Twitter 上分享”按钮时,我编写了预检查代码。我刚刚添加了这段代码。
-
嗯,我遇到了同样的错误,但我意识到我的代码在 accountStore 被初始化之前被调用了。这显然不是你的问题。也许在 performRequestWithHandler 处设置一个断点并检查变量的内容?
-
我已经检查了一切,但没有成功。
标签: ios objective-c cocoa twitter slrequest