为什么我的 UIImageView 替换了第二个？答案

【问题标题】：Why does my UIImageView replace the second one?为什么我的 UIImageView 替换了第二个？
【发布时间】：2017-12-03 20:18:22
【问题描述】：

我有两个用于两个不同 UIImageView 的插座，当我选择第一个时，它会出现在第一个 Image View 上，但是当我选择第二个 Image 时，即使它连接到第二个 ImageView，它也会替换第一个 Image View。这是我的“选择图像”按钮的代码。

@IBOutlet weak var myImageView1: UIImageView!
@IBOutlet weak var myImageView2: UIImageView!


@IBAction func pickImage1(_ sender: Any) {

    let image = UIImagePickerController()
    image.delegate = self
    image.sourceType = UIImagePickerControllerSourceType.photoLibrary
    image.allowsEditing = false

    self.present(image, animated: true)
}

//Add didFinishPickingMediaWithInfo here
func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]) {
    if let image = info[UIImagePickerControllerOriginalImage] as? UIImage {
        myImageView1.image = image
    }
    else {
        //error
    }
    self.dismiss(animated: true, completion: nil)
}


@IBAction func pickImage2(_ sender: Any) {
    let image2 = UIImagePickerController()
    image2.delegate = self
    image2.sourceType = UIImagePickerControllerSourceType.photoLibrary
    image2.allowsEditing = false

    self.present(image2, animated: true)
}

//Add didFinishPickingMediaWithInfo here
func imagePickerController2(_ picker2: UIImagePickerController, didFinishPickingMediaWithInfo2 info2: [String : Any]) {
    if let image2 = info2[UIImagePickerControllerOriginalImage] as? UIImage {
        myImageView2.image = image2
    }
    else {
        //error
    }
    self.dismiss(animated: true, completion: nil)

}

【问题讨论】：

如果您重命名委托方法，就像您在第二次调用中所做的那样，它将不再被识别或调用。
@BJHStudios 您对问题原因的解释应该在您的回答中。
@rmaddy 完成，谢谢。

标签： ios swift swift3 uiimageview imageview

【解决方案1】：

试试这个代码。所以你需要一个标志来记住点击了哪个图像视图，然后根据它设置图像。

var selected = 1

@IBAction func pickImage1(_ sender: Any) {

    let image = UIImagePickerController()
    image.delegate = self
    image.sourceType = UIImagePickerControllerSourceType.photoLibrary
    image.allowsEditing = false
    selected = 1

    self.present(image, animated: true)
}

//Add didFinishPickingMediaWithInfo here
func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]) {
    if let image = info[UIImagePickerControllerOriginalImage] as? UIImage {
        if selected == 1 {
            myImageView1.image = image
        } else {
            myImageView2.image = image
        }
    }
    else {
        //error
    }
    self.dismiss(animated: true, completion: nil)
}


@IBAction func pickImage2(_ sender: Any) {
    let image2 = UIImagePickerController()
    image2.delegate = self
    image2.sourceType = UIImagePickerControllerSourceType.photoLibrary
    image2.allowsEditing = false
    selected = 2

    self.present(image2, animated: true)
}

向前看，当您有多个图像视图时，您可以使用另一种方法来避免到处复制代码。

首先，为每个图像视图添加一个唯一标签。避免使用 0，因为默认标签是 0。因此您将看到带有标签的图像视图，例如 1 到 4。

对所有图像视图调用相同的方法，以便通过单击其中任何一个来触发此功能

let tapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(imageTapped(tapGestureRecognizer:)))
imageView.addGestureRecognizer(tapGestureRecognizer)

处理程序看起来像这样

func imageTapped(tapGestureRecognizer: UITapGestureRecognizer)
{
    let image = UIImagePickerController()
    image.delegate = self
    image.sourceType = UIImagePickerControllerSourceType.photoLibrary
    image.allowsEditing = false
    let tappedImage = tapGestureRecognizer.view as! UIImageView
    selected = tappedImage.tag
    self.present(image, animated: true)
}

终于在图片挑选委托中

func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]) {
    if let image = info[UIImagePickerControllerOriginalImage] as? UIImage {
        if let imageView = self.view.viewWithTag(selected) as? UIImageView {
            imageView.image = image
        }
    }
    else {
        //error
    }
    self.dismiss(animated: true, completion: nil)
}

【讨论】：

虽然这是一个正确的解决方案，但为了使这个答案更好，请在 OP 的代码中解释问题。
与其将selected 属性添加为Int，不如添加var selectedImageView: UIImageView? 属性。在两个 pickImageX 方法中设置它。然后在delegate中，直接将选中的图片赋值给属性。不需要if。
嘿，如果我想添加额外的图像，如 3 或 4，我该怎么办？
@Riccardo 更新

【解决方案2】：

问题是您重命名了委托方法。如果你这样做，该方法将不会被识别或调用。

所选答案的另一个选项是扩展 UIImageView 并使其遵守 UIImagePickerControllerDelegate / UINavigationControllerDelegate。

extension UIImageView: UIImagePickerControllerDelegate, UINavigationControllerDelegate {

     public func imagePickerController(_ picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [String : Any]) {
          guard let selectedImage = info[UIImagePickerControllerOriginalImage] as? UIImage else {
             //handle error
             return
         }

         image = selectedImage
         picker.presentingViewController?.dismiss(animated: true)
     }

     func presentImagePicker(from viewController: UIViewController) {
         let picker = UIImagePickerController()
         picker.delegate = self
         picker.sourceType = .photoLibrary
         picker.allowsEditing = false

         viewController.present(picker, animated: true)
     }
}

这很好，因为您可以使用一行代码为您的应用中的任何UIImageView 启动图像选择器，如下所示：

@IBAction func pickImage1(_ sender: UIButton) {
    myImageView1.presentImagePicker(from: self)
}

【讨论】：