ios - 如何解决此错误 : Could not cast value of type '__NSCFString' (0x10354a248) to 'UIImage' (0x104c42b48)

Question

当我运行代码并单击 segue 以描述 View Controller 时，在调试中出现错误。无法将类型“__NSCFString”(0x10354a248)的值转换为“UIImage”(0x104c42b48)。在主视图 Controller 错误中得到了这个 Thread 1: signal SIGABRT

MainViewController.swift

导入 UIKit进口 Alamofire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
    }
}

Answer 1

这一行是罪魁祸首:

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

您正在尝试将 String 转换为 UIImage。

一个解决方案是向 foods[indexPath.row] 添加另一个名为“photos”的键，并确保在分配它时它是一个 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 foods 数组的类型替换为 [Food]。这应该会在代码中提供更多安全性，并且您不必在每次想要拍照时都强制转换。