Question

我试过搜索但找不到答案。我写了一个应用程序，我正在尝试分享内容到Facebook。基本上我想分享一个URL，也许是一个引用或标题。

我一直收到一个名为'reserved'的错误，但我不确定它是什么意思或如何修复它。任何帮助都会很棒！

func fbClick() {

    let content = LinkShareContent(url: URL(string: "www.google.com")!)
    showShareDialog(content, mode: .native)

}

func showShareDialog<C: ContentProtocol> (_ content: C, mode: ShareDialogMode = .automatic) {
    let dialog = ShareDialog(content: content)
    dialog.presentingViewController = self
    dialog.mode = mode

    do {
        try dialog.show()
    } catch (let error) {
        self.view.makeToast("Invalid share content. Failed to present share dialog with error \(error)", duration: 3.0, position: .top)
    }
}

Answer 1

想出来。

这一行...

let content = LinkShareContent(url: URL(string: "www.google.com")!)

应该是这样的......

let content = LinkShareContent(url: NSURL(string: "https://www.google.com")! as URL)

或者像这样

let content = LinkShareContent(url: NSURL(string: "https://www.google.com")! as URL, quote: quote)

Answer 2

使用reserved时出现相同的VideoShareContent错误。花了5个小时找到问题，终于找到了。真的希望有人发现这也很有帮助。

解决方案：当您从url委托方法的info param检索视频的UIImagePickerController时，请确保使用密钥{{1 }}

示例：

"UIImagePickerControllerReferenceURL"

其他信息：最初我没有使用此密钥“UIImagePickerControllerReferenceURL”。为什么：它是deprecated。根据Apple的说法，你应该使用UIImagePickerControllerPHAsset代替。但是那里的url也会返回func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String: Any]) { picker.dismiss(animated: true) if let videoURL = info["UIImagePickerControllerReferenceURL"] as? URL { let video = Video(url: videoURL) let content = VideoShareContent(video: video) do { try ShareDialog.show(from: self, content: content) } catch { print(error) } } }错误。另一种尝试是使用键reserved，但它也没有成功。

IOS高级错误'保留'被退回

2 个答案: