使用 AlamofireObjectMapper/ObjectMapper 从数组 objects 获取值 (swift - iOS)

Question

我是这个映射器的新手，太困惑了。我有一个 API 请求，给定一个标题，API return 这个：

{
Response = True;
Search =     (
            {
        Poster = "https://images-na.ssl-images-amazon.com/images/M/MV5BMjAxODQ2MzkyMV5BMl5BanBnXkFtZTgwNjU3MTE5OTE@._V1_SX300.jpg";
        Title = ARQ;
        Type = movie;
        Year = 2016;
        imdbID = tt5640450;
    },
            {
        Poster = "N/A";
        Title = Arq;
        Type = movie;
        Year = 2011;
        imdbID = tt2141601;
    },
            {
        Poster = "N/A";
        Title = "A.R.Q.";
        Type = movie;
        Year = 2015;
        imdbID = tt3829612;
    }
);
totalResults = 3;

}

所以我为此结果创建了一个可映射的class：

class SearchResponse: Mappable {
    var isSuccess  : String?
    var searchArray: [Movie]?
    var searchCount: String?

    required init?(map: Map) {
    }

    func mapping(map: Map) {
        isSuccess   <- map["Response"]
        searchArray <- map["Search"]
        searchCount <- map["totalResults"]
    }
}

class Movie: Mappable {

    var posterURL  : String?
    var title      : String?
    var runtime    : String?
    var director   : String?
    var actors     : String?
    var genre      : String?
    var plot       : String?
    var production : String?
    var year       : String?
    var imdbID     : String?
    var imdbRating : String?

    required init?(map: Map) {

    }

    func mapping(map: Map) {
        posterURL  <- map["Poster"]
        title      <- map["Title"]
        runtime    <- map["Runtime"]
        director   <- map["Director"]
        actors     <- map["Actors"]
        genre      <- map["Genre"]
        plot       <- map["Plot"]
        production <- map["Production"]
        year       <- map["Year"]
        imdbID     <- map["imdbID"]
        imdbRating <- map["imdbRating"]
    }
}

Question: I mapped this movie class like this, but for the search by title I'll only have 4 of this attributes. But for the next search I'll need all of them. Is that right? Or should I create two separate classes to deal with each kind of response?

好的！我正在我的 SearchTableViewController 上显示此搜索的结果。现在我想显示这部电影的更多细节（之前响应中 "Search" 数组的任何电影）。为此，API 提供了另一种搜索类型，即按 imdbID 搜索。因此，我在我的 SearchTableViewController 上创建了一个 segue 以获取此 ID 并传递给我的 MovieViewController（将显示这些详细信息的视图）：

let searchSegue = "segueFromSearch"

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    let searchIndex = tableView.indexPathForSelectedRow?.row
    let movie = movies?[searchIndex!]
    let selectedImdbID = movie?.imdbID

    print("|Table View Controler| Segue. IMDB_ID: \(String(describing: selectedImdbID))")

    if segue.identifier == searchSegue {
        if let destination = segue.destination as? MovieViewController {

            destination.imdbID = selectedImdbID!

            print("|Table View Controler| Inside of if let. Debug print: I get til here. imdbID = \(selectedImdbID!)")
        }

    }
}

我对这个 API 请求的函数是：

//The movieSearched variable is the text typed on my searchBar
let URL = "https://www.omdbapi.com/?s=\(movieSearched)&type=movie"
    Alamofire.request(URL).responseObject{ (response: DataResponse<SearchResponse>) in
        print("response is: \(response)")

        switch response.result {
        case .success(let value):
            let searchResponse = value
            self.movies = (searchResponse.searchArray)
            self.searchTableView.reloadData()


        case .failure(let error):
            let alert = UIAlertController(title: "Error", message: "Error 4xx / 5xx: \(error)", preferredStyle: UIAlertControllerStyle.alert)
            alert.addAction(UIAlertAction(title: "OK", style: UIAlertActionStyle.default, handler: nil))
            self.present(alert, animated: true, completion: nil)
        }

    }

好的，鉴于我所拥有的这个概述，让我们谈谈我的问题...

当我按 ID 搜索时，Json 响应是：

{
    Actors = "Robbie Amell, Rachael Taylor, Shaun Benson, Gray Powell";
    Awards = "N/A";
    BoxOffice = "N/A";
    Country = "USA, Canada";
    DVD = "16 Sep 2016";
    Director = "Tony Elliott";
    Genre = "Sci-Fi, Thriller";
    Language = English;
    Metascore = "N/A";
    Plot = "Trapped in a lab and stuck in a time loop, a disoriented couple fends off masked raiders while harboring a new energy source that could save humanity.";
    Poster = "https://images-na.ssl-images-amazon.com/images/M/MV5BMjAxODQ2MzkyMV5BMl5BanBnXkFtZTgwNjU3MTE5OTE@._V1_SX300.jpg";
    Production = Netflix;
    Rated = "N/A";
    Ratings =     (
        {
            Source = "Internet Movie Database";
            Value = "6.4/10";
        },
        {
            Source = "Rotten Tomatoes";
            Value = "60%";
        }
    );
    Released = "16 Sep 2016";
    Response = True;
    Runtime = "88 min";
    Title = ARQ;
    Type = movie;
    Website = "N/A";
    Writer = "Tony Elliott";
    Year = 2016;
    imdbID = tt5640450;
    imdbRating = "6.4";
    imdbVotes = "17,481";
}

我的问题

我通过 ID 发出了这个 alamofire 搜索请求：

func getMovieById() {

    let URL = "https://www.omdbapi.com/?i=\(String(describing: imdbID!)))"

    Alamofire.request(URL).responseObject{ (response: DataResponse<SearchResponse>) in
        print("|MovieController| Response is: \(response)")

        let Result = response.result.value
        print("Result for search by id: \(String(describing: Result!.searchArray))")

        // Have to get the values here, right?

    }
}

显然，我在这里没有得到我想要的数据。所以...

Questions:

如何使用可映射的 classes 获取 Json["Search"] 的值？
我必须更改现有的 classes 吗？如果是，如何以及为什么？

层层叠叠看得我眼花缭乱。另外，我是 swift 的初学者，我是第一次使用这个 ObjectMapper。抱歉这里有这么多代码，但我想我必须解释一下我的场景。

Answer 1

您必须使用属性的正确 数据类型 映射每个属性。您的响应中的对象之一包含 布尔值 值（例如 "Response"），但您将其声明为字符串。我们必须准确匹配属性的数据类型，否则对象将为 nil 并且不会被映射。

此外 按 ID 搜索 响应与您的映射器不匹配 class。

let Result = response.result.value 是错误的。 response.result.value 会给你 SearchResponse 对象。

底线

你必须先把映射部分弄好。任何不匹配的类型都不会被映射。使用响应对象将为您提供包含所有映射的整个对象，而不是 JSON 显然。所以应该是：let movie = response.result.value。然后你访问电影的属性，比如 ex. movie.actors

使用 AlamofireObjectMapper/ObjectMapper 从数组 objects 获取值 (swift - iOS)

Get value from array objects with AlamofireObjectMapper/ObjectMapper (swift - iOS)

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