2016-11-04 84 views
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我正在尝试使用UIActivityView为我的应用程序创建共享按钮。当我按下应用程序中的按钮时,出现错误。代码有什么问题?UIActivityView按钮错误

@IBAction func shareButton(_ sender: UIButton) { 

     let texttoshare = "Sharing" 
     let URLtoshare = NSURL(string: "www.google.com") 
     let objectsToShare:NSArray = [texttoshare, URLtoshare!] 
     let activityVC = UIActivityViewController(activityItems: objectsToShare as! [Any], applicationActivities: nil) 

     self.present(activityVC, animated: true, completion: nil) 

      } 
    } 

错误:是线程1:信号SIGABRT

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请将错误添加到您的问题。 –

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尝试更改'NSURL(字符串:...'到URL(字符串:...),并删除':NSArray' –

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@LucianoRodríguez仍然是同样的错误... – Phillip

回答

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试试这个。至少,它的工作对我来说没有errores ..

let texttoshare = "Sharing" 
let URLtoshare = URL(string: "www.google.com") 
let activityVC = UIActivityViewController(activityItems: [texttoshare, URLtoshare!], applicationActivities: nil) 
self.present(activityVC, animated: true, completion: nil) 

如果你想使用UIActivityViewController为ipad公司,你需要为activityVC指定popoverPresentationController?.sourceView。例如:

let texttoshare = "Sharing" 
let URLtoshare = URL(string: "www.google.com") 
let activityVC = UIActivityViewController(activityItems: [texttoshare, URLtoshare!], applicationActivities: nil) 
activityVC.popoverPresentationController?.sourceView = button... 
self.present(activityVC, animated: true, completion: nil) 
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你把什么“按钮... “?我仍然得到相同的错误 – Phillip

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任何你想要的...例如shareButton IBOutlet ...或者self.view,如果你愿意的话,测试几个视图,看看有什么适合你的需求...... –