如何解决此错误：无法将类型“__NSCFString”（0x10354a248）的值转换为“UIImage”（0x104c42b48）

Posted 2023-02-24

【中文标题】如何解决此错误：无法将类型“__NSCFString”（0x10354a248）的值转换为“UIImage”（0x104c42b48）

【英文标题】：How to Solve This Error : Could not cast value of type '__NSCFString' (0x10354a248) to 'UIImage' (0x104c42b48)

【发布时间】：2019-02-19 21:35:45

【问题描述】：

当我运行代码并单击 segue 以描述视图控制器时，调试中出现错误。无法将类型“__NSCFString”（0x10354a248）的值转换为“UIImage”（0x104c42b48）。 在主视图控制器错误得到这个线程1：信号SIGABRT

MainViewController.swift

导入 UIKit 进口阿拉莫火 导入 AlamofireImage 导入 SDWebImage

类 MainViewController: UIViewController, UITableViewDataSource, UITableViewDelegate

@IBOutlet weak var tableView: UITableView?
var foods: [[String: Any]] = [[String: Any]]()


override func viewDidLoad() 
    super.viewDidLoad()
    Alamofire.request("https://api.myjson.com/bins/1bnsyj").responseJSON  (response) in
        if let responseValue = response.result.value as! [String: Any]? 
            print(responseValue)
            if let responseFoods = responseValue["items"] as! [[String: Any]]? 
                self.foods = responseFoods
                self.tableView?.reloadData()

            
        

        else 
            print("error : \(String(describing: response.result.error))")
        
    


    // Do any additional setup after loading the view.
 
// MARK: - UITableViewDataSource & UITableViewDelegate

func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int 
    return foods.count


func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell 
    let cell = tableView.dequeueReusableCell(withIdentifier: "FoodTableViewCell") as! FoodTableViewCell
    if foods.count > 0 
        let eachFood = foods[indexPath.row]
        cell.lblFoodName?.text = (eachFood["name"] as? String) ?? ""
        cell.lblDescription?.text = (eachFood["description"] as? String) ?? ""

        let url = NSURL(string: self.foods[indexPath.row]["photoUrl"]! as! String)
        cell.imageViewFood?.af_setImage(withURL: url! as URL, placeholderImage: nil, filter: nil,runImageTransitionIfCached: true, completion: nil)

    

    return cell

func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) 
    let vc = storyboard?.instantiateViewController(withIdentifier: "DetailsViewController") as! DetailsViewController
    vc.foodnameseg = foods[indexPath.row]["name"] as! String
    vc.descriptionsegue = foods[indexPath.row]["description"] as! String
    vc.imagefoodsegue = foods[indexPath.row]["photoUrl"] as! UIImage
    self.navigationController?.pushViewController(vc, animated: true)

DetailsViewController.swift

import UIKit

class DetailsViewController: UIViewController 
    var foodnameseg : String = ""
    var descriptionsegue : String = ""
    var imagefoodsegue = UIImage()


    @IBOutlet weak var imagefood: UIImageView!
    @IBOutlet weak var lblNameSegue: UILabel!
    @IBOutlet weak var descriptionSegue: UITextView!
        override func viewDidLoad() 
        super.viewDidLoad()

         lblNameSegue.text = foodnameseg
         descriptionSegue.text = descriptionsegue
         imagefood.image = imagefoodsegue
    

【参考方案1】：

这行是罪魁祸首：

vc.imagefoodsegue = foods[indexPath.row]["photoUrl"] as! UIImage

您正在尝试将String 转换为UIImage。

对此的一种解决方案是向 foods[indexPath.row] 添加另一个名为“photos”的键，并在分配时确保它是 UIImage。

或者你可以像这样制作一个食物结构：

struct Food

  var name: String
  var photoURL: URL
  var photo: UIImage?

将您的 foods 数组类型替换为 [Food]。这应该会在代码中提供更高的安全性，并且您不必每次想要获取照片时都强制投射。