将图像从 UIImagePickerController 传递到另一个 ViewController答案

【问题标题】：Pass image from UIImagePickerController to another ViewController将图像从 UIImagePickerController 传递到另一个 ViewController
【发布时间】：2014-04-02 09:20:42
【问题描述】：

我正在阅读很多关于此的帖子，但我无法解决我的问题... 我正在尝试使用 Picker 将图像从一个 ViewController 发送到另一个 ViewController，但图像没有出现...

我有 2 个 VC：

HomeViewController.h：

#import <UIKit/UIKit.h>
#import "PhotoViewController.h"

@interface QuizTypeViewController : UIViewController <UIImagePickerControllerDelegate, UINavigationControllerDelegate>

- (IBAction)photo:(id)sender;

@end

HomeViewController.m（我得到了正确的图像，我将只发布 segue 代码）

- (void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info {

UIImage *chosenImage = info[UIImagePickerControllerEditedImage];

PhotoViewController *controller = [PhotoViewController new];
controller.imageView.image = chosenImage;

[picker dismissViewControllerAnimated:YES completion:NULL];
[self performSegueWithIdentifier:@"viewPhoto" sender:self];

}

PhotoViewController.h

#import <UIKit/UIKit.h>

@interface PhotoViewController : UIViewController
@property (strong, nonatomic) IBOutlet UIImageView *imageView;
@end

PhotoViewController.m - 没什么...

我做错了什么？我不知道...

【问题讨论】：

你合成了imageview吗？
@Ramdy 在带有LLVM 的 Xcode 5 中，不需要合成等等。
那么，我们不需要使用 synthesize 吗？ @rckoenes
@Ramdy 不，如果您使用 LLVM，现在是强制性的。您不再需要在.m 中添加@synthesize

标签： ios objective-c cocoa-touch uiimagepickercontroller

【解决方案1】：

你不应该new PhotoViewController。当你打电话时

[self performSegueWithIdentifier:@"viewPhoto" sender:self];

将自动为您创建一个 PhotoViewController 实例。您应该做的是将所选图像传递给它。并在您的PhotoViewController 的一些方法（例如：viewDidLoad）中显示它。

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    PhotoViewController *photoViewController = segue.destinationViewController;
    photoViewController.image = self.chosenImage;
}

【讨论】：

在 viewDidLoad 我必须插入这个：“self.imageView.image = self.image;”

【解决方案2】：

您在 imagePickerController:didFinishPickingMediaWithInfo: 中创建了新的 PhotoViewController，但您没有在视图层次结构中推送/呈现它，因此它将被关闭。最好的方法是在 performSegueWithIdentifier:sender 方法中将图像作为参数传递：

- (void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info {

UIImage *chosenImage = info[UIImagePickerControllerEditedImage];
_tmp = chosenImage;

[picker dismissViewControllerAnimated:YES completion:NULL];
[self performSegueWithIdentifier:@"viewPhoto" sender: chosenImage];

}

并在 prepareForSegue:segue: 方法中从发送者获取图像并将其传递给目标视图控制器：

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    // TODO: check segue identifier
    PhotoViewController *vc = (PhotoViewController*)segue.destinationViewController;
    // Get the image 
    UIImage *img = (UIImage*)sender
    // Pass image to the new view controller.
    vc.imageView.image = img;
    //It can failed because your image view can not be created
    // You should use @property for UIImage, pass img to image and in view did load
    //assign imageView.image = image
}

【讨论】：

【解决方案3】：

[self performSegueWithIdentifier:@"viewPhoto" sender:self]; 会自己创建一个新的 ViewController 实例。

如果您希望它显示您的实例，您需要使用 [self presentViewController :viewController animated:YES] 或类似名称显示

【讨论】：

【解决方案4】：

 - (void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info 
{

  UIImage *chosenImage = info[UIImagePickerControllerEditedImage];
 _tmp = chosenImage;

  PhotoViewController *controller    = [self.storyboard instantiateViewControllerWithIdentifier:@"viewPhoto"];
controller.imageView.image = chosenImage;

[picker dismissViewControllerAnimated:YES completion:NULL];
[self presentViewController:controlle animated:YES completion:nil];
}

【讨论】：

【解决方案5】：

HomeViewController.m

PhotoViewController *controller = [PhotoViewController new];
controller.image = chosenImage;

PhotoViewController.h

@property (weak, nonatomic) IBOutlet UIImage *image;
@property (strong, nonatomic) IBOutlet UIImageView *imageView;

PhotoViewController.m

imageView.image = image;

【讨论】：