从UIImagePickerController传递图像到另一个ViewController

我正在阅读很多关于此的帖子，但我无法解决我的问题... 我试图发送一个图像从一个ViewController与选择器...到另一个ViewController但图像没有出现......从UIImagePickerController传递图像到另一个ViewController

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（我得到正确的图像，我要去发布只是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......

我在做什么错了？我不知道...

来源

2014-04-02 pekpon

你有没有合成的ImageView的？ – Ramdy

@Ramdy在带有LLVM的Xcode 5中，不需要合成等等。 – rckoenes

那么，我们不需要使用合成？@rckoenes – Ramdy

你不应该new一个PhotoViewController。当你调用

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

一个PhotoViewController实例将自动为您创建。你应该做的是将选定的图像传递给它。并在你的PhotoViewController的某些方法（例如：viewDidLoad）中显示它。

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

来源

2014-04-02 09:26:57 sunkehappy

中添加'@ synthesize'，我必须插入：“self.imageView.image = self.image;” – pekpon

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

如果你想让它显示你的情况下，你需要[self presentViewController :viewController animated:YES]或类似

来源

2014-04-02 09:28:15 HellGate

您在imagePickerController创建新PhotoViewController显示：didFinishPickingMediaWithInfo：但你不考虑层次推动/目前它，所以它会被驳回。最好的方式是如在performSegueWithIdentifier参数通图像：发送方法：

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

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

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

}

和在prepareForSegue：赛格瑞：方法从发送端获得的图像并传递给目的地视图控制器：

- (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 
}

来源

2014-04-02 09:28:34 Greg

- (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]; 
}

来源

2014-04-02 09:32:17 Rose

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;

来源

2014-04-02 09:36:29 Ramdy

从UIImagePickerController传递图像到另一个ViewController

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