如何使UIImageView正确的框架大小？

Question

I'm trying to take a picture using imagePickerController and show the image in a UIImageView in another ViewController.

- (void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info {
    [picker dismissViewControllerAnimated:YES completion:NULL];
    PAWWallPostCreateViewController *wallPostCreateViewController = [[PAWWallPostCreateViewController alloc] initWithNibName:@"PAWWallPostCreateViewController" bundle:nil];

    wallPostCreateViewController.myImage = info[UIImagePickerControllerEditedImage];
    wallPostCreateViewController.imageView = [[UIImageView alloc] initWithImage:wallPostCreateViewController.myImage];
    wallPostCreateViewController.imageView.contentMode = UIViewContentModeScaleAspectFit;

    NSLog(@"%@",wallPostCreateViewController.imageView); // Output below
    [self.navigationController presentViewController:wallPostCreateViewController animated:YES completion:nil];
}

The output in NSLog for the imageView is:

<UIImageView: 0x18f695b0; frame = (0 0; 640 640); opaque = NO; userInteractionEnabled = NO; layer = <CALayer: 0x18f14240>>

The frame is much too big for my actual UIImageView which is set to (60 227; 200 200)

I think this is the reason why I can't see UIImageView at all when I run my app. How do I fix this?

EDIT:

In PAWWallPostCreateViewController.h

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

I linked imageView to my .xib file

Also is anyone familiar with autoresize = RM+BM; Because usually when I have used imagePickerController to put the UIImage in a UIImageView of the same view controller my UIImageView data has a autoresize = RM+BM; attribute so the code I want usually looks like

<UIImageView: 0x1a2b4b30; frame = (60 227; 200 200); opaque = NO; autoresize = RM+BM; userInteractionEnabled = NO; layer = <CALayer: 0x1a2b4bb0>>

But that's now what I'm getting in this case

Answer 1

只需在ViewDidLoad（）的第二个视图控制器下调用此行

self.imageView.contentMode = UIViewContentModeScaleAspectFit;