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  1. POHide/Show iPhone Camera Iris/Shutter animation
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    copied!<p>I am not able to Hide the iphone Camera shutter opening animation for my app. I am using UIImagePickerController to access iphone camera and using my own overlay controllers. Is there a way to remove the initial shutter(also known as Iris) animation as the camera starts. Thank You</p> <p>[EDIT]</p> <p>For those who wants to know the way to change the camera iris animation.</p> <p>The below function is called before the camera iris animation starts.</p> <pre><code>- (void)navigationController:(UINavigationController *)navigationController willShowViewController:(UIViewController *)viewController animated:(BOOL)animated { // Here is were I make the camera preview fit the entire screen. // This might violate the "don't change the view hierarchy"-rule. // So I am not sure if it is valid for App Store commitment. // However, the NSLogs are used to // figure out which subview is the actual Camera Preview which turns out // to be the PLPreviewView. (uncomment to se the printouts). // Change it's size to fit the entire screen (and scale it accordingly // to avoid distorted image NSLog(@"WillShowViewController called..."); NSLog(@"VC:view:subviews\n %@\n\n", [[viewController view] subviews]); NSLog(@"VC:view:PLCameraView:subviews\n %@\n\n", [[[[viewController view] subviews] objectAtIndex: 0] subviews]); NSLog(@"VC:view:PLCameraView:PLPreviewView:subviews\n %@\n\n", [[[[[[viewController view] subviews] objectAtIndex: 0] subviews] objectAtIndex: 0] subviews]); NSLog(@"VC:view:PLCameraView:PLCropOverLay:subviews\n %@\n\n", [[[[[[viewController view] subviews] objectAtIndex: 0] subviews] objectAtIndex: 1] subviews]); NSLog(@"VC:view:PLCameraView:UIImageView:subviews\n %@\n\n", [[[[[[viewController view] subviews] objectAtIndex: 0] subviews] objectAtIndex: 2] subviews]); } </code></pre> <p>In the above function you can go through each layer by using the normal NSMuatableArray syntax like objectAtIndex</p> <p>hope this might help you.</p> <p>Regards,</p> <p>Ankur</p>
 

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