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我一直在使用一个食谱应用程序工作一段时间，您可以在其中使用勺子 API 搜索食谱。我已成功将 JSON 数据存储在一个数组中，但现在我不知道当用户搜索某些内容时如何将数据获取到 tableview？到目前为止，这是我的代码。

我如何获取 JSON 文件并读取数据：

Class APIRequest() {

static let instance = APIRequest()

var query:String = ""
var url: String = ""
var testString: String = ""

func setQuery(_ query: String) -> Void {
    self.query = query
}

func getReturn(completed: @escaping (Result<Response, Error>) -> Void) {
    // TODO MAJOR!
    // Expressions are not allowed at top level will occur otherwise
    //

    let headers = [
        "x-rapidapi-host": "spoonacular-recipe-food-nutrition-v1.p.rapidapi.com",
        "x-rapidapi-key": "e44daac5e0mshc682df24497a89fp1c4513jsn7067934f0b9b"
    ]

    var request = URLRequest(url: URL(string: "https://spoonacular-recipe-food-nutrition-v1.p.rapidapi.com/recipes/search?number=2&query=\(query)")!,
                             cachePolicy: .useProtocolCachePolicy,
                             timeoutInterval: 10.0)
    request.httpMethod = "GET"
    request.allHTTPHeaderFields = headers

    let session = URLSession.shared
    let dataTask = session.dataTask(with: request) { (data, response, error) -> Void in
        if let error = error { completed(.failure(error));  return }
        do {
            let result = try JSONDecoder().decode(Response.self, from: data!)
            completed(.success(result))

        } catch {
            completed(.failure(error))
        }
    }
    dataTask.resume()
}

我的食谱课：

class Recipe: Codable {
let id: Int
let image: String
let title: String
let readyInMinutes: Int

init(id: Int, image: String = "Default", title: String = "Title", readyInMinutes: Int) {
    self.id = id
    self.image = image
    self.title = title
    self.readyInMinutes = readyInMinutes
}
}

配方处理器：

class RecipeHandler {

static let instance = RecipeHandler()

var allRecipeResults: [Recipe] = []
}

然后我有这个我不确定的响应结构：

struct Response: Codable {
   struct Result: Codable {
       let id: Int
       let title: String
       let image: String
       let readyInMinutes: Int
   }
   let results: [Result]
   }

还有我的 RecipeViewController：

class RecipeViewController: UITableViewController {

@IBOutlet var recipeTable: UITableView!

override func viewDidLoad() {
    super.viewDidLoad()

    recipeTable.delegate = self
    recipeTable.dataSource = self


}


}


//MARK: - Add cell
extension RecipeViewController {

override func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int{
    return RecipeHandler.instance.allRecipeResults.count
    }

override func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
    if let cell = tableView.dequeueReusableCell(withIdentifier: "RecipeCell") as? RecipeTableViewCell{
        let recipeItem = RecipeHandler.instance.allRecipeResults[indexPath.row]
        cell.recipeImage.image = UIImage(named: recipeItem.image)
        cell.recipeTitle.text = recipeItem.title
        cell.recipeTime.text = "\(recipeItem.readyInMinutes)"
        return cell
    }
    return UITableViewCell()
    }
    }

//MARK: - SearchBarSearchButtinClicked
extension RecipeViewController: UISearchBarDelegate{
func searchBarSearchButtonClicked(_ searchBar: UISearchBar) {
    guard let searchBarText = searchBar.text else {return}
    let request = APIRequest.instance
    request.query = searchBarText
    request.getReturn { result in
        switch result {
            case .success(let result):
                print(result.results)
                //Don't know what to do here?

            case .failure(let error):
                print(error)
        }
    }

}
}
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1 回答 1

0

您似乎有正确的想法...您将RecipeHandler单例用作简单的数据模型，并尝试从tableView:cellForRowAtIndexPath:. 您只是从不将任何数据放入模型中，因此没有什么可以输出的。

        case .success(let result):
            print(result.results)
            //Don't know what to do here?

.success在您提出请求后，您的处理程序是否会被调用？结果会打印出来吗？如果不是，那么您需要在响应处理和解码中寻找问题。否则，将Results 转换为食谱并将它们添加到您的模型中。

于 2019-12-03T22:33:31.047 回答