使用popvivew控制器时，如何将Uiimage从一个视图控制器传递到另一个视图控制器？

Question

I Have Two View controllers. 我有两个View控制器。

First UIViewController contain one UIImage, when I click "crop" button, go to second view controller I pass the image by push view controller, here I'am cropping my UIImage. 第一个UIViewController包含一个UIImage，当我单击“裁剪”按钮时，转到第二个视图控制器，我通过推式视图控制器传递图像，这里我裁剪了UIImage。
In second UIViewController after cropping my image, clicking "done" button, I need to pass the cropped image to first view. 裁剪图像后，在第二个UIViewController中，单击“完成”按钮，我需要将裁剪的图像传递到第一视图。

Problem is when I use pop-view controller, image is not passing. 问题是当我使用弹出视图控制器时，图像未通过。

Answer 1

1)Make a UIIMAGE *variable in first view controllers header. 1）在第一个视图控制器标头中设置UIIMAGE * variable。

2)set the uiimage variable from second view controller before popping. 2）在弹出之前从第二个视图控制器设置uiimage变量。

3)POP the second view controller. 3）POP第二个视图控制器。

If this is not working make a singleton class consisting of uiimage variable 如果这不起作用，则使一个由uiimage变量组成的单例类

Answer 2

Try like this. 尝试这样。 First view controller should have a property 第一视图控制器应具有一个属性

@property (assign)UIImage *croppedImage;

Second view controller should have 第二个视图控制器应具有

@property (assign)FirstViewController *first;

Before popping set 弹出集之前

first.croppedImage = self.image;