türünün değerini dönüştürülemez, yardıma mı ihtiyacınız '(NSURLRequest !, NSHTTPURLResponse !, UIImage!)

Question

Im

türünün değerini dönüştürülemez diyen bir hata alıyorum (NSURLRequest !, NSHTTPURLResponse !, UIImage!) - > Beklenen argüman türüne 'Void' ((NSURLRequest, NSHTTPURLResponse ?, UIImage) -> Void)? İşte

kodu

import UIKit 

class RoutineCell: UITableViewCell { 

@IBOutlet weak var bannerImageView: UIImageView! 
@IBOutlet weak var routineTitleLabel: UILabel! 
@IBOutlet weak var shortDescriptionLabel: UILabel! 

var routine: Routine! { 
    didSet { 
     routineTitleLabel.text = routine.title 
     shortDescriptionLabel.text = routine.shortDescription 
     bannerImageView.setImageWithURL(NSURL(string: routine.bannerImageUrl)!) 
     bannerImageView.setImageWithURLRequest(NSURLRequest(URL: NSURL(string: routine.bannerImageUrl)!), placeholderImage: nil, success: { (request: NSURLRequest!, response: NSHTTPURLResponse!, image: UIImage!) -> Void in 
      self.bannerImageView.alpha = 0.0 
      self.bannerImageView.image = image 
      UIView.animateWithDuration(0.2, animations: {() -> Void in 
       self.bannerImageView.alpha = 1.0 
      }) 
     }, failure: nil) 
    } 
} 

override func awakeFromNib() { 
    self.selectionStyle = UITableViewCellSelectionStyle.None 
    super.awakeFromNib() 
} 

override func prepareForReuse() { 
    bannerImageView.image = nil 
} 

override func setSelected(selected: Bool, animated: Bool) { 
    super.setSelected(selected, animated: animated) 

    // Configure the view for the selected state 
} 

} 

sayesinde

olduğunu

Answer 1

bannerImageView.setImageWithURLRequest geri arama için Beyanınız aşağıdaki

(request: NSURLRequest, response: NSHTTPURLResponse?, image: UIImage) 

kullanmak gerekir

(request: NSURLRequest!, response: NSHTTPURLResponse!, image: UIImage!) 

yanlıştır