2016-07-29 128 views
0

我试图共享一个URL,以便在Safari或任何Web浏览器中打开URL,但无法这样做。我已经发送到我的MAC和MAC打开它作为文本文件,然后通过点击,我要去safari,但在iPhone的情况下我努力打开它。在NSURL的情况下,我的应用程序崩溃与一些例外。这里是我的代码无法通过蓝牙共享URL(UIActivityViewController)

NSString *[email protected]"www.google.com"; 
UIActivityViewController *controller = [[UIActivityViewController alloc] initWithActivityItems:@[url] applicationActivities:nil]; 

// Exclude all activities except AirDrop. 
NSArray *excludedActivities = @[UIActivityTypePostToTwitter, UIActivityTypePostToFacebook, 
           UIActivityTypePostToWeibo, 
           UIActivityTypeMessage, UIActivityTypeMail, 
           UIActivityTypePrint, UIActivityTypeCopyToPasteboard, 
           UIActivityTypeAssignToContact, UIActivityTypeSaveToCameraRoll, 
           UIActivityTypeAddToReadingList, UIActivityTypePostToFlickr, 
           UIActivityTypePostToVimeo, UIActivityTypePostToTencentWeibo]; 
controller.excludedActivityTypes = excludedActivities; 

// Present the controller 
[self presentViewController:controller animated:YES completion:nil]; 

回答

1

改变这一行:

NSString *[email protected]"www.google.com"; 

到:

NSURL *url= [NSURL URLWithString:@"https://www.google.com"]; 

如果这不适合你,因为崩溃的很可能是个问题,在其他地方,但使崩溃报告可用,也许我可以帮助。