Question

我正在实施转发功能，但在POST后我一直收到错误的网址错误。这是我的代码：

SLRequest *twitterRequest = [SLRequest requestForServiceType:SLServiceTypeTwitter requestMethod:SLRequestMethodPOST URL:[NSURL URLWithString:@"https://api.twitter.com/1.1/statuses/retweet/%@.json"] parameters:[NSDictionary dictionaryWithObject:tweetId forKey:@"id"]];

[twitterRequest performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) {
  dispatch_async(dispatch_get_main_queue(), ^{

    if ([urlResponse statusCode] == 429) {
        NSLog(@"Rate limit reached");
        return;
    }

    if (error) {
        NSLog(@"Error: %@", error.localizedDescription);
        return;
    }

  });
}];

有什么想法吗？我错过了什么吗？谢谢！

Answer 1

我的坏。我在创建请求时忘了传递字符串“...％@。json”。

即

SLRequest *twitterRequest = [SLRequest requestForServiceType:SLServiceTypeTwitter requestMethod:SLRequestMethodPOST URL:[NSURL URLWithString:[NSString stringWithFormat:@"https://api.twitter.com/1.1/statuses/retweet/%@.json",@"490794578250059776"]] parameters:nil];

Answer 2

这是

的内容

   SLRequest *twitterRequest = [SLRequest requestForServiceType:SLServiceTypeTwitter requestMethod:SLRequestMethodPOST URL:[NSURL URLWithString:@"https://api.twitter.com/1.1/statuses/retweet/%@.json"] parameters [NSDictionary dictionaryWithObject:tweetId forKey:@"id"]];

试试这个

 SLRequest *twitterRequest = [SLRequest requestForServiceType:SLServiceTypeTwitter requestMethod:SLRequestMethodPOST URL:[NSURL URLWithString:@"https://api.twitter.com/1.1/statuses/retweet/%@.json"] parameters:[NSDictionary dictionaryWithObject:tweetId forKey:@"status"]];

有一件事你不能一次又一次地重新发布同一个消息，它会被视为垃圾邮件并且看到这个link可能你可能会错过一些东西

Answer 3

您应该在您的请求中附加一个帐户。

// create a request
NSURL *url = [NSURL URLWithString:@"https://api.twitter.com/1.1/statuses/user_timeline.json"];

SLRequest *request = [SLRequest requestForServiceType:SLServiceTypeTwitter
                                        requestMethod:SLRequestMethodGET
                                                  URL:url
                                           parameters:@{@"screen_name":@"nst021"}];

// attach an account to the request
NSArray *twitterAccounts = [accountStore accountsWithAccountType:twitterAccountType];
[request setAccount:[twitterAccounts lastObject]];

// execute the request
[request performRequestWithHandler:^(NSData *responseData,
                                     NSHTTPURLResponse *urlResponse,
                                     NSError *error) {
    // caution, you're on an arbitrary queue here...
}

使用SLRequest进行iOS转发会返回错误的网址错误

3 个答案: