将 UIImageView 从主视图移动到另一个

Question

我的代码可以很好地从图库中选择图片并将其显示在同一视图中，现在遇到的问题是，当单击 "next" 按钮时，将选择的 UIImageView 转移到下一个 activity

这是打开图库的代码

@IBAction func gallery(sender: AnyObject) {
    if UIImagePickerController.availableMediaTypesForSourceType(.PhotoLibrary) != nil {
        picker.allowsEditing = false
        picker.sourceType = UIImagePickerControllerSourceType.PhotoLibrary
        presentViewController(picker, animated: true, completion: nil)
    }

}

这是在同一视图中显示图像的代码

func imagePickerController(picker: UIImagePickerController, didFinishPickingMediaWithInfo info: [NSObject : AnyObject]) {
        var chosenImage = info[UIImagePickerControllerOriginalImage] as! UIImage 
        imageChosen.contentMode = .ScaleAspectFit 
        imageChosen.image = chosenImage 
        dismissViewControllerAnimated(true, completion: nil) 

    }

现在，在 UIImageView 保存在 imageChosen 之后，这里是用于将该图像传递到下一个视图的代码

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
    var pass:postView = segue.destinationViewController as! postView
    if(segue.identifier == "next"){
        pass.imgv.image = imageChosen.image
    }
}

导致程序崩溃的这行代码

pass.imgv.image = imageChosen.image

第二个view中的imgv是这样声明的

@IBOutlet weak var imgv: UIImageView!

我哪里做错了，请指导我

Answer 1

您无法在第二个视图控制器中直接访问 UIImageView。而是在第二个视图控制器中创建变量 UIImage 并将您选择的 UIImage 分配给它。稍后在第二个视图控制器的 viewdidload 中设置 UIImageView.

第一次观看

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
    let pass:postView = segue.destinationViewController as! postView
    if(segue.identifier == "next"){
        pass.tempImage = imageChosen.image
    }
}

第二个视图

@IBOutlet weak var imgv: UIImageView!
var tempImage:UIImage!

override func viewDidLoad() {
    super.viewDidLoad()
    self.imgv.image=tempImage
}

Answer 2

您正在为图像视图使用弱引用并试图在其他控制器中获取它。做如下 .

@IBOutlet 强变量 imgv: UIImageView

其次，您获取视图控制器的方式错误...尝试如下

if segue.identifier == "ShowCounterSegue"
{
    if let destinationVC = segue.destinationViewController as? OtherViewController{
        destinationVC.numberToDisplay = counter
     }
 }

Answer 3

在 PostView 中，只需创建变量 var img: UIImage!
将这一行 pass.imgv.image = imageChosen.image 替换为 pass.img = imageChosen.image
在PostView的viewDidLoad()中，添加这一行，imgv.image = image

我们不能从firstView设置secondView的IBOutlets属性，因为ViewController的IBOutlets会在viewDidLoad()[=38=执行后获取内存]

因此您必须创建适当的数据变量以将数据传递给 IBOutlet。

这是示例代码。

class FirstViewController: UIViewController { @IBOutlet weak var imageViewInput: UIImageView! override func viewDidLoad() { super.viewDidLoad() } override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) { //check the condition for your segue.identifier, if any var destination : NewViewContrller! = segue.destinationViewController as NewViewContrller ; if let image = imageViewInput?.image { destination.outputImage = image; } } } class NewViewContrller: UIViewController { @IBOutlet weak var outputImageView: UIImageView! var outputImage: UIImage! override func viewDidLoad() { super.viewDidLoad() // Do any additional setup after loading the view, typically from a nib. if let image = outputImage { outputImageView.image = image; } } }

Answer 4

您不能在尚未呈现的视图中设置数据。因此，将图像传递给第二个视图，并在第二个视图的 viewDidLoad

中将该图像设置为 ImageView

  override func prepareForSegue(segue: (UIStoryboardSegue!), sender: AnyObject!) {
        if segue.identifier == "next" {
            var pass:second = segue.destinationViewController as! second
            pass.currentImage=myImageView.image;
        }
    }

//第二个视图

class second: UIViewController {

    @IBOutlet weak var tempImgView: UIImageView!
    var currentImage:UIImage!

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view.
        if((currentImage) != nil){
            tempImgView.image=currentImage;
        }
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
    }

}

将 UIImageView 从主视图移动到另一个

moving the UIImageView from the main view to another

pass-data

uiimageview

ios

swift