iOS 5推特框架:在没有用户输入和确认的情况下进行推文(模式视图控制器)

Question

本质上，我想要的是，一旦用户允许访问他们的Twitter帐户，应用程序就可以在UITableView中发布用户选择的任何内容。理想情况下，我想在iOS5中使用Twitter框架，但我遇到的主要问题是推特的模式视图控制器。这是可选的吗？没有它可以发推特吗？如果没有，你建议我怎么做？

谢谢！

Answer 1

如果没有它，推特肯定是可能的，以下是iOS 5应用程序的生产版本。如果用户还没有注册帐户，它甚至会将其带到首选项的必要部分。

- (void)postToTwitter
{
    // Create an account store object.
    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) {
            // Get the list of Twitter accounts.
            NSArray *accountsArray = [accountStore accountsWithAccountType:accountType];

            if ([accountsArray count] > 0) {
                // Grab the initial Twitter account to tweet from.
                ACAccount *twitterAccount = [accountsArray objectAtIndex:0];
                TWRequest *postRequest = nil;
                postRequest = [[TWRequest alloc] initWithURL:[NSURL URLWithString:@"http://api.twitter.com/1/statuses/update.json"] parameters:[NSDictionary dictionaryWithObject:[self stringToPost] forKey:@"status"] requestMethod:TWRequestMethodPOST];

                // Set the account used to post the tweet.
                [postRequest setAccount:twitterAccount];
                dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^(void) {
                    [postRequest performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) {
                        dispatch_async(dispatch_get_main_queue(), ^(void) {
                            if ([urlResponse statusCode] == 200) {
                                Alert(0, nil, @"Tweet Successful", @"Ok", nil);
                            }else {
                                Alert(0, nil, @"Tweet failed", @"Ok", nil);
                            }
                        });
                    }];
                });
            }
            else
            {
                [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=TWITTER"]];
            }
        }
    }];
}

Answer 2

这将是一个使用SLRequest而不是TWRequest的更新版本，TWRequest在iOS 6中已被弃用。请注意，这需要将社会和账户框架添加到您的项目中。。。

- (void) postToTwitterInBackground {
    // Create an account store object.
    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 options:nil completion:^(BOOL granted, NSError *error) {
        if(granted) {
            // Get the list of Twitter accounts.
            NSArray *accountsArray = [accountStore accountsWithAccountType:accountType];
            if ([accountsArray count] > 0) {
                // Grab the initial Twitter account to tweet from.
                ACAccount *twitterAccount = [accountsArray objectAtIndex:0];
                SLRequest *postRequest = nil;
                // Post Text
                NSDictionary *message = @{@"status": @"Tweeting from my iOS app!"};
                // URL
                NSURL *requestURL = [NSURL URLWithString:@"https://api.twitter.com/1.1/statuses/update.json"];
                // Request
                postRequest = [SLRequest requestForServiceType:SLServiceTypeTwitter requestMethod:SLRequestMethodPOST URL:requestURL parameters:message];
                // Set Account
                postRequest.account = twitterAccount;
                // Post
                [postRequest performRequestWithHandler:^(NSData *responseData, NSHTTPURLResponse *urlResponse, NSError *error) {
                     NSLog(@"Twitter HTTP response: %i", [urlResponse statusCode]);
                 }];
            }
        }
    }];
}

Answer 3

更新：Fabric by Twitter中的TwitterKit非常方便，如果用户试图在你的应用程序中发推文时，你打算从你的推特应用程序中发布，那么这可能是一个不错的选择。

（是的，这种方法将允许你在没有任何对话框或确认的情况下发布到推特上）。

TwitterKit将处理权限部分，并使用TWTRAPIClient通过Twitter rest API执行推文。

 //Needs to performed once in order to get permissions from the user to post via your twitter app.
[[Twitter sharedInstance]logInWithCompletion:^(TWTRSession *session, NSError *error) {
    //Session details can be obtained here
    //Get an instance of the TWTRAPIClient from the Twitter shared instance. (This is created using the credentials which was used to initialize twitter, the first time) 
    TWTRAPIClient *client = [[Twitter sharedInstance]APIClient];
    //Build the request that you want to launch using the API and the text to be tweeted.
    NSURLRequest *tweetRequest = [client URLRequestWithMethod:@"POST" URL:@"https://api.twitter.com/1.1/statuses/update.json" parameters:[NSDictionary dictionaryWithObjectsAndKeys:@"TEXT TO BE TWEETED", @"status", nil] error:&error];
   //Perform this whenever you need to perform the tweet (REST API call)
   [client sendTwitterRequest:tweetRequest completion:^(NSURLResponse *response, NSData *data, NSError *connectionError) {
   //Check for the response and update UI according if necessary.            
   }];
}];

希望这能有所帮助。

Answer 4

由于多次更改，接受的答案不再有效。这一款适用于iOS 10、Swift 3和Twitter API 1.1版。

**更新**

这个答案已经更新，因为上一个答案依赖于一个不推荐使用的Twitter端点。

import Social
import Accounts
func postToTwitter() {
    let accountStore = ACAccountStore()
    let accountType = accountStore.accountType(withAccountTypeIdentifier: ACAccountTypeIdentifierTwitter)
    accountStore.requestAccessToAccounts(with: accountType, options: nil) { (granted, error) in
        if granted, let accounts = accountStore.accounts(with: accountType) {
            // This will default to the first account if they have more than one
            if let account = accounts.first as? ACAccount {
                let requestURL = URL(string: "https://api.twitter.com/1.1/statuses/update.json")
                let parameters = ["status" : "Tweet tweet"]
                guard let request = SLRequest(forServiceType: SLServiceTypeTwitter, requestMethod: .POST, url: requestURL, parameters: parameters) else { return }
                request.account = account
                request.perform(handler: { (data, response, error) in
                    // Check to see if tweet was successful
                })
            } else {
                // User does not have an available Twitter account
            }
        }
    }
}

这是正在使用的API。