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社区首页 >问答首页 >iOS用织物发送tweet而没有组合tweet

iOS用织物发送tweet而没有组合tweet
EN

Stack Overflow用户
提问于 2015-02-11 14:04:24
回答 2查看 2.3K关注 0票数 2

我想发送推特,但我不想使用撰写推特。

代码语言:javascript
复制
TWTRComposer *composer = [[TWTRComposer alloc] init];

[composer setText:@"just setting up my Fabric"];
[composer setImage:[UIImage imageNamed:@"fabric"]];

[composer showWithCompletion:^(TWTRComposerResult result) {
    if (result == TWTRComposerResultCancelled) {
        NSLog(@"Tweet composition cancelled");
    }
    else {
        NSLog(@"Sending Tweet!");
    }
}];

https://dev.twitter.com/twitter-kit/ios/compose这个方法显示一些弹出屏幕,我想发送tweet即时我怎么能做到这与面料api?这种类型有可能吗?

此代码部分不适用于织物。

代码语言:javascript
复制
 ACAccountStore *accountStore = [[ACAccountStore alloc] init];
    ACAccountType *accountType = [accountStore accountTypeWithAccountTypeIdentifier:ACAccountTypeIdentifierTwitter];

    [accountStore requestAccessToAccountsWithType:accountType withCompletionHandler:^(BOOL granted, NSError *error)
     {
         if(granted)
         {
             NSArray *accountsArray = [accountStore accountsWithAccountType:accountType];

             if ([accountsArray count] > 0)
             {
                 ACAccount *twitterAccount = [accountsArray objectAtIndex:0];

                 TWRequest *postRequest = [[TWRequest alloc] initWithURL:[NSURL URLWithString:@"https://upload.twitter.com/1/statuses/update_with_media.json"] parameters:[NSDictionary dictionaryWithObject:self.textViewOutlet.text forKey:@"status"] requestMethod:TWRequestMethodPOST];

                 [postRequest addMultiPartData:UIImagePNGRepresentation(image) withName:@"media" type:@"multipart/png"];
                 [postRequest setAccount:twitterAccount];

                 [postRequest performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error)
                  {
                      //show status after done
                      NSString *output = [NSString stringWithFormat:@"HTTP response status: %i", [urlResponse statusCode]];
                      NSLog(@"Twiter post status : %@", output);
                  }];
             }
         }
     }];
EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2015-02-12 18:27:51

https://github.com/nst/STTwitter

我使用了这个API,比fabric容易:)

代码语言:javascript
复制
[_twitter postStatusUpdate:@"tweet text" inReplyToStatusID:nil latitude:nil longitude:nil placeID:nil displayCoordinates:nil trimUser:nil successBlock:nil errorBlock:nil];

很容易在没有作曲家的情况下发推文可能是这个人帮助了某人..。

票数 2
EN

Stack Overflow用户

发布于 2016-03-29 07:22:03

试试这个:

代码语言:javascript
复制
func tweet(userId: String) {
        let client = TWTRAPIClient(userID: userId)

        let error: NSErrorPointer = NSErrorPointer()
        let url: String = "https://api.twitter.com/1.1/statuses/update.json"
        let message: [NSObject : AnyObject] = [
            "status" : "Sample Tweet Tweet!"
        ]

        let preparedRequest: NSURLRequest = client.URLRequestWithMethod("POST", URL: url, parameters: message, error: error)

        client.sendTwitterRequest(preparedRequest) { (response, data, jsonError) -> Void in
            do {
                let json = try NSJSONSerialization.JSONObjectWithData(data!, options: []) as! [String:AnyObject]
                NSLog("%@", json)
                print("Tweet post!")
            } catch {
                print("json error: \(error)")
            }
        }
    }

您必须首先获得用户的userId。如果您正在使用手动按钮注册,请使用以下命令:

代码语言:javascript
复制
func tweet(userId: String) {
        let client = TWTRAPIClient(userID: userId)

        let error: NSErrorPointer = NSErrorPointer()
        let url: String = "https://api.twitter.com/1.1/statuses/update.json"
        let message: [NSObject : AnyObject] = [
            "status" : "Sample Tweet Tweet!"
        ]

        let preparedRequest: NSURLRequest = client.URLRequestWithMethod("POST", URL: url, parameters: message, error: error)

        client.sendTwitterRequest(preparedRequest) { (response, data, jsonError) -> Void in
            do {
                let json = try NSJSONSerialization.JSONObjectWithData(data!, options: []) as! [String:AnyObject]
                NSLog("%@", json)
                print("Tweet post!")
            } catch {
                print("json error: \(error)")
            }
        }
    }

很抱歉我用Swift写了这段代码。

票数 3
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/28456336

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