How to Solve This Error : Could not cast value of type '__NSCFString' (0x10354a248) to 'UIImage' (0x104c42b48)
How to Solve This Error : Could not cast value of type '__NSCFString' (0x10354a248) to 'UIImage' (0x104c42b48)
当我 运行 代码并单击 segue 到描述视图控制器时,在调试中出现错误。无法将类型 '__NSCFString' (0x10354a248) 的值转换为 'UIImage' (0x104c42b48).
在主视图控制器错误得到这个 Thread 1: signal SIGABRT
MainViewController.swift
导入 UIKit
进口 Alamofire
导入 AlamofireImage
导入 SDWebImage
class 主视图控制器: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
}
}
这一行是罪魁祸首:
vc.imagefoodsegue = foods[indexPath.row]["photoUrl"] as! UIImage
您正在尝试将 String
转换为 UIImage
。
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 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的安全性,并且您不必在每次想要拍照时都强制转换。
当我 运行 代码并单击 segue 到描述视图控制器时,在调试中出现错误。无法将类型 '__NSCFString' (0x10354a248) 的值转换为 'UIImage' (0x104c42b48). 在主视图控制器错误得到这个 Thread 1: signal SIGABRT
MainViewController.swift
导入 UIKit 进口 Alamofire 导入 AlamofireImage 导入 SDWebImage
class 主视图控制器: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
}
}
这一行是罪魁祸首:
vc.imagefoodsegue = foods[indexPath.row]["photoUrl"] as! UIImage
您正在尝试将 String
转换为 UIImage
。
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 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的安全性,并且您不必在每次想要拍照时都强制转换。