Swift

Question

我有 2 个 UIImageView，我在其中添加了一个 UITapGestureRecognizer，用于打开用户的相机。一旦用户选择了图像，为了访问相机而点击的 UIImageView 应该设置为相机图像。

这是我的 openCamera 方法：

   @objc func openCamera(sender: ImagePlaceHolderView) {
       let picker = UIImagePickerController()
        picker.allowsEditing = true
        picker.delegate = self
        picker.sourceType = .camera 
        present(picker, animated: true)
    }

我的 UIImageView 声明如下：

var left = UIImageView()
var right = UIImageView()
left = createImagePlaceholder()
right = createImagePlaceholder()

private func createImagePlaceholder() -> UIImageView {
    let placeholder = ImagePlaceHolderView(frame: CGRect(x: 0, y: 0, width: 90, height: 150))
    placeholder.isUserInteractionEnabled = true
    return placeholder
}

然后是我的委托方法：

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {
    guard let image = info[.editedImage] as? UIImage else { return }
    left.image = image  // Here I want to set the sender's image to image
    dismiss(animated: true, completion: nil)
}

我正在努力弄清楚如何将点击的 UIImageView 的图像设置为来自相机的新图像。我试图将发件人传递给 openCamera 方法，但不确定如何提取它，因为发件人实际上似乎是 UITapGestureRecognizer 而不是被点击的图像。

Answer 1

您必须存储对 sender 的引用，并使用它在委托方法中设置图像。

var imagePlaceholderView: ImagePlaceHolderView? // set the reference

@objc func openCamera(sender: UITapGestureRecognizer) {
    imagePlaceholderView = sender.view as? ImagePlaceHolderView
    //...
}

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [UIImagePickerController.InfoKey : Any]) {
    guard let image = info[.editedImage] as? UIImage else { return }
    imagePlaceholderView?.image = image
    dismiss(animated: true, completion: nil)
}

Swift - 获取 UIImagePickerController 的发送者

Swift - get sender of UIImagePickerController

xcode

uiimageview

uiimagepickercontroller

uitapgesturerecognizer