swift 共享按钮在iPhone上工作完美，但在iPad上崩溃

iezvtpos 于 2022-12-10 发布在 Swift

关注(0)|答案(3)|浏览(149)

我试图添加一个按钮，以便在Twitter，Facebook等分享一些句子。它都适用于所有iPhone型号，但模拟器崩溃与iPad。
这是我的代码：

@IBAction func shareButton(sender: AnyObject) {
    frase = labelFrases.text!
    autor = labelAutores.text!

    var myShare = "\(frase) - \(autor)"
    
    let activityVC: UIActivityViewController = UIActivityViewController(activityItems: [myShare], applicationActivities: nil)

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

这就是错误：
由于未捕获的异常“NSGenericException”，正在终止应用程序，原因：'UI弹出演示控制器（〈_UI警报控制器操作表常规演示控制器：0x7c0f9190〉）应在显示之前设置非空的sourceView或barButtonItem
我该怎么解决呢？

swift

来源：https://stackoverflow.com/questions/31506081/sharing-button-works-perfectly-on-iphone-but-crash-on-ipad

3条答案

按热度按时间

qeeaahzv1#

对于iPad（iOS〉8.0），您需要设置popoverPresentationController：

//check ipad
if (UIDevice.currentDevice().userInterfaceIdiom == UIUserInterfaceIdiom.Pad)
{
    //ios > 8.0
    if ( activityVC.respondsToSelector(Selector("popoverPresentationController"))){
        activityVC.popoverPresentationController?.sourceView = super.view
    }
}

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

更多信息，请访问：UIActivityViewController crashing on iOS 8 iPads

赞(0）回复(0）举报 2022-12-10

but5z9lq2#

请改为执行以下操作，以便Swift 5在iPad和iPhone上都能使用共享按钮：

@IBAction func shareButton(sender: UIButton) { {
    let itemToShare = ["Some Text goes here"]
    let avc = UIActivityViewController(activityItems: itemToShare, applicationActivities: nil)
    
    //Apps to be excluded sharing to
    avc.excludedActivityTypes = [
        UIActivityType.print,
        UIActivityType.addToReadingList
    ]
    // Check if user is on iPad and present popover
    if UIDevice.current.userInterfaceIdiom == .pad {
        if avc.responds(to: #selector(getter: UIViewController.popoverPresentationController)) {
            avc.popoverPresentationController?.barButtonItem = sender
        }
    }
    // Present share activityView on regular iPhone
    self.present(avc, animated: true, completion: nil)
}

希望这对你有帮助！

赞(0）回复(0）举报 2022-12-10

nhn9ugyo3#

稍微修改版本，使其工作在任何按钮，iPad和iPhone。
Xcode 13.4.1（快速移动5.6）

let itemToShare = ["Some Text goes here"]
    let avc = UIActivityViewController(activityItems: itemToShare, applicationActivities: nil)
    
    //Apps to be excluded sharing to
    avc.excludedActivityTypes = [
        UIActivity.ActivityType.print,
        UIActivity.ActivityType.addToReadingList
    ]
    // Check if user is on iPad and present popover
    if UIDevice.current.userInterfaceIdiom == .pad {
        if avc.responds(to: #selector(getter: UIViewController.popoverPresentationController)) {
            avc.popoverPresentationController?.sourceView = sender as? UIView
        }
    }
    // Present share activityView on regular iPhone
    self.present(avc, animated: true, completion: nil)

赞(0）回复(0）举报 2022-12-10

我来回答

swift 共享按钮在iPhone上工作完美，但在iPad上崩溃

3条答案

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