为什么我的UIImageView取代了第二个？

Posted 2021-04-18

我有两个不同的UIImageViews的插座，当我选择第一个它将出现在第一个图像视图时，但当我选择第二个图像时，它取代了第一个图像视图，即使它连接到第二个ImageView。这是我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)

}

答案

试试这个代码。因此，您需要一个标记来记住单击哪个图像视图，然后根据该标记设置图像。

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)
}

继续前进，当您有多个图像视图时，可以使用另一种方法来避免在任何地方复制代码。

首先，为每个图像视图添加唯一标记。避免使用0，因为默认标记为0.因此，您将拥有标记为1到4的图像视图。

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)
}

最后在图片选择委托

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)
}

另一答案

问题是您已重命名委托方法。如果这样做，将无法识别或调用该方法。

选定答案的另一个选择是扩展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为应用程序中的任何UIImageView启动图像选择器，如下所示：

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