当我运行代码并单击segue to description视图控制器时，在Debugging中出现错误。无法将类型'__NSCFString'（0x10354a248）的值强制转换为'UIImage'（0x104c42b48）。在主视图控制器错误中得到此线程1：信号SIGABRT
MainViewController.swift
导入UIKit导入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
}
}
这是罪魁祸首：
vc.imagefoodsegue = foods[indexPath.row]["photoUrl"] as! UIImage
您正在尝试将转换String为UIImage。
解决此问题的一种方法是在foods [indexPath.row]中添加另一个称为“照片”的密钥，并确保UIImage在分配该密钥时使用它。
或者，您可以像这样制作食物结构：
struct Food
{
var name: String
var photoURL: URL
var photo: UIImage?
}
将foods数组的类型替换为[Food]。这样可以在代码中提供更大的安全性，并且您不必每次想要获取照片时都强制播报。
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