iOS6 - 社交框架 - SLComposeViewController如何回退到iOS5的TWTweetComposeViewController？

Question

我有一块推特共享代码,适用于iOS6但我需要应用程序才能很好地回退到iOS5 ......

它看起来像这样:

- (void) shareOnTwitter
{
    if([SLComposeViewController instanceMethodForSelector:@selector(isAvailableForServiceType)] != nil)
    {
        if ([SLComposeViewController isAvailableForServiceType:SLServiceTypeTwitter])
        {
            NSLog(@"twitter available");
            SLComposeViewController *composeViewController = [SLComposeViewController composeViewControllerForServiceType:SLServiceTypeTwitter];
            [composeViewController setInitialText:self.sharingText];
            [self.sharingController presentViewController:composeViewController animated:YES completion:nil];
        }
        else
        {
            NSLog(@"twitter not available!");
        }
    }
    else
    {
        // SLComposeViewController not available, this is most likely <iOS6, what to do here?
    }
}

那么,我如何在iOS5中很好地退回(我假设我需要TWTweetComposeViewController)以便我也可以在iOS5中使用本机twitter？

编辑:最后我仍然懒得回退到TWTweetComposeViewController,所以我决定简单回归这个序列:iOS6原生推文 - >安装推特应用程序 - >网址.这是我放在一起的功能,希望它可以帮助某人:

+(BOOL)isSocialFrameworkAvailable
{
     // whether the iOS6 Social framework is available?
    return NSClassFromString(@"SLComposeViewController") != nil;
}

- (void) shareOnTwitterWithText:(NSString*)text andURL:(NSString*)url andImageName:(NSString*)imageName
{
    // prepare the message to be shared
    NSString *combineMessage = [NSString stringWithFormat:@"%@ %@", text, url];
    NSString *escapedMessage = [combineMessage stringByAddingPercentEscapesUsingEncoding:NSASCIIStringEncoding];
    NSString *appURL = [NSString stringWithFormat:@"twitter://post?message=%@", escapedMessage];

    if([SocialManager isSocialFrameworkAvailable] && [SLComposeViewController isAvailableForServiceType:SLServiceTypeTwitter])
    {
        // user has setup the iOS6 twitter account

        SLComposeViewController *composeViewController = [SLComposeViewController composeViewControllerForServiceType:SLServiceTypeTwitter];
        [composeViewController setInitialText:text];
        if([UIImage imageNamed:imageName])
        {
            [composeViewController addImage:[UIImage imageNamed:imageName]];
        }
        if(url)
        {
            [composeViewController addURL:[NSURL URLWithString:url]];
        }
        [self.sharingController presentViewController:composeViewController animated:YES completion:nil];
    }
    else
    {
        // else, we have to fallback to app or browser
        if([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:appURL]])
        {
            // twitter app available!
            [[UIApplication sharedApplication] openURL:[NSURL URLWithString:appURL]];
        }
        else
        {
            // worse come to worse, open twitter page in browser
            NSString *web = [NSString stringWithFormat:@"https://twitter.com/intent/tweet?text=%@", escapedMessage];
            [[UIApplication sharedApplication] openURL:[NSURL URLWithString:web]];
        }
    }
}

Answer 1

您需要弱化Twitter和社交框架的链接,并检查您的代码是否有框架可用.弱链接是这样完成的:

• 在XCode中单击您的项目,选择Target,Build Phases,然后选择Link Binary with Libraries
• 如果要在iOS 5.x或更早版本上运行此应用程序,请确保将Social.framework设置为Optional
• 如果要在iOS 4.x或更早版本上运行此选项,请确保将Twitter.framework设置为Optional

我喜欢创建简单的类函数来确定哪些框架可用.这可能看起来像这样:

+(BOOL)isTwitterAvailable {
   return NSClassFromString(@"TWTweetComposeViewController") != nil;
}

+(BOOL)isSocialAvailable {
    return NSClassFromString(@"SLComposeViewController") != nil;
}

您的"推文"代码可能如下所示:

if ([SomeClass isSocialAvailable]) {
   // code to tweet with SLComposeViewController
} else if ([SomeClass isTwitterAvailable]) {
   // code to tweet with TWTweetComposeViewController
} else {
   // Twitter not available, or open a url like https://twitter.com/intent/tweet?text=tweet%20text
}

Answer 2

不确定这些运行时操作有多昂贵,但这样做没有任何害处,因为在应用程序运行时此状态不会发生变化:

+ (BOOL)isTwitterAvailable
{
    static BOOL available;
    static dispatch_once_t onceToken;
    dispatch_once(&onceToken, ^{
        available = NSClassFromString(@"TWTweetComposeViewController") != nil;        
    });
    return available;
}
+ (BOOL)isSocialAvailable
{
    static BOOL available;
    static dispatch_once_t onceToken;
    dispatch_once(&onceToken, ^{
        available = NSClassFromString(@"SLComposeViewController") != nil;        
    });
    return available;
}