无法使用“UIImagePickerController.InfoKey”类型的索引为“[String : Any]”类型的值下标答案

【问题标题】：Cannot subscript a value of type '[String : Any]' with an index of type 'UIImagePickerController.InfoKey'无法使用“UIImagePickerController.InfoKey”类型的索引为“[String : Any]”类型的值下标
【发布时间】：2018-12-22 19:59:11
【问题描述】：

我正在使用Apple's Swift iOS Tutorial。这是抛出一个错误，

不能使用“UIImagePickerController.InfoKey”类型的索引为“[String : Any]”类型的值下标

他们定义的函数如下。

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]) {

    // The info dictionary may contain multiple representations of the image. You want to use the original.
    guard let selectedImage = info[UIImagePickerControllerOriginalImage] as? UIImage else {
        fatalError("Expected a dictionary containing an image, but was provided the following: \(info)")
    }

    // Set photoImageView to display the selected image.
    photoImageView.image = selectedImage

    // Dismiss the picker.
    dismiss(animated: true, completion: nil)
}

我使用的是 Xcode 10.0 beta 3，其中包括 Swift 4.2。

我想了解如何遍历文档以了解可能已更改或损坏的内容。

【问题讨论】：

我注意到本教程后面的另一行也抛出了一个错误 - button.addTarget(self, action: #selector(RatingControl.ratingButtonTapped(button:)), for: .touchUpInside)。如果其他人有同样的问题——我可以通过在课程行的开头添加@objcMembers 来解决它，就像这样：@objcMembers class RatingControl: UIStackView {。见hackingwithswift.com/example-code/language/…

标签： ios swift uiimagepickercontroller

【解决方案1】：

Swift 4.2 中方法的签名发生了变化

func imagePickerController(_ picker: UIImagePickerController, 
  didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any])

你必须写

guard let selectedImage = info[.originalImage] as? UIImage else {
    fatalError("Expected a dictionary containing an image, but was provided the following: \(info)")
}

您可以通过阅读documentation 或注释掉整个方法、重新输入前几个字符并使用代码完成来自己找出这些术语的变化。

【讨论】：

很好，它已经清除了 IDE 中的错误。就像我刚开始做 iOS 并遵循相同教程的 OP。
有人收到类似“'UIImage”的错误吗？在“as?”中不能转换为“UIImage”” ?

【解决方案2】：

我也在学习相同的教程，更新后的代码如下所示：

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {

    // The info dictionary may contain multiple representations of the image. You want to use the original.
    guard let selectedImage = info[.originalImage] as? UIImage else {
        fatalError("Expected a dictionary containing an image, but was provided the following: \(info)")
    }

    // Set photoImageView to display the selected image.
    photoImageView.image = selectedImage

    // Dismiss the picker.
    dismiss(animated: true, completion: nil)
}

【讨论】：

【解决方案3】：

在 Swift 4.2 中对方法进行了一些更改。

首先初始化这两个变量：

var imagePicker = UIImagePickerController()
var pickedImageProduct = UIImage()

extension yourViewController: UIImagePickerControllerDelegate,UINavigationControllerDelegate{
//create an IBAction to access Camera
    @IBAction func accessCameraBtn(_ sender: UIButton) {
        imagePicker.delegate = self
        imagePicker.sourceType = UIImagePickerController.SourceType.camera
        self.present(imagePicker, animated: true, completion: nil)
    }
//create an IBAction to access Gallery
    @IBAction func accessGalleryBtn(_ sender: UIButton) {
        imagePicker.delegate = self
        imagePicker.sourceType = UIImagePickerController.SourceType.photoLibrary
        self.present(imagePicker, animated: true, completion: nil)
    }
//Final step put this Delegate    
    func imagePickerController(_ picker: UIImagePickerController,
                               didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {

        guard let selectedImage = info[.originalImage] as? UIImage else {
           Print("Error: \(info)")
        }

            pickedImageProduct = selectedImage

        dismiss(animated: true, completion: nil)
     }
    func imagePickerControllerDidCancel(_ picker: UIImagePickerController) {

        dismiss(animated: true, completion: nil)
    }

}

【讨论】：

方法改得更快了，我想4.0 甚至可能更早。

【解决方案4】：

在 Swift 4.2 中，您可以将函数编写为：

    func photoLib()
    {
        //opens Photo Library, call the function in a @IBAction func
        let myPickerController = UIImagePickerController()
        myPickerController.delegate = self;
        myPickerController.sourceType = 
        UIImagePickerController.SourceType.photoLibrary
        myPickerController.allowsEditing = true
        present(myPickerController, animated: true)
    }


    func imagePickerController(_ picker: UIImagePickerController, 
    didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {
    picker.dismiss(animated: true)

    guard let image = info[.editedImage] as? UIImage else {
        print("No image found")
        photoLabel.text = "Photo Not Found"
        return
    }
}

【讨论】：

【解决方案5】：

这样使用，

guard let selectedImage = info[UIImagePickerController.InfoKey.originalImage.rawValue] as? UIImage else {
        fatalError("Expected a dictionary containing an image, but was provided the following: \(info)")
}

【讨论】：

【解决方案6】：

在 Swift 4 和 5 中是这样的：

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {

    guard let selectedImage = info[.editedImage] as? UIImage else {
        fatalError("Expected a dictionary containing an image, but was provided the following: \(info)")
    }

    self.myImageView.image = selectedImage

    picker.dismiss(animated: true, completion: nil)
}

【讨论】：

Swift 4 和 5 没有区别

【解决方案7】：

斯威夫特 5

在 swift 4 或 5 的最新版本中，委托方法如下所示，它应该可以工作

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {
    // Code here
}

并使用，

guard let selectedImage = info[.editedImage] as? UIImage else {

}



func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {
    guard let selectedImage = info[.editedImage] as? UIImage else {

    }
}

【讨论】：

【解决方案8】：

我也在关注教程。如下更改后它正在工作。

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {

    // The info dictionary may contain multiple representations of the image. You want to use the original.
    guard let selectedImage = info[.originalImage] as? UIImage
        else {
        fatalError("Expected a dictionary containing an image, but was provided the following: \(info)")
    }

    // Set photoImageView to display the selected image.
    photoImageView.image = selectedImage

    // Dismiss the picker.
    dismiss(animated: true, completion: nil)
}

我正在使用 XCode 11 和 Swift 5。

【讨论】：