Google Books API解析

时间:2017-08-08 15:17:27

标签: ios swift google-api google-books

所以我试图解析Google Books API的回应。我想获得标题,描述,thumbnailUrl,作者和发布的数据。这是问题所在:

func getBooksFrom(completion: @escaping (Result<[[String: AnyObject]]>) -> Void) {
    URLSession.shared.dataTask(with: url) { (data, response, error) in
        guard error == nil else { return completion(.Error(error!.localizedDescription)) }
        guard let data = data else { return
            completion(.Error(error!.localizedDescription)) }
        do {
            if let json = try JSONSerialization.jsonObject(with: data, options: [.mutableContainers]) as? [String: AnyObject] {
                if let items = json["items"] as? [[String: AnyObject]] {
                    DispatchQueue.main.async {
                        completion(.Succes(items))
                    }
                }
            }
        } catch let error {
            print(error.localizedDescription)
            return completion(.Error(error.localizedDescription))
        }
    }.resume()
}

在ViewDidLoad的View Controller中我有

let service = ApiService()
    service.getBooksFrom { (result) in
        switch result {
        case .Succes(let data):
            self.parseData(array: data)
        case .Error(let message):
            self.showAlertWith(title: "Error", and: message)
        }
    }

这样很简单的解析,但...... 当我想将项目映射到Book Object时,我必须:

func parseData(_ data: [[String: AnyObject]]) -> [Book]{
    for item in data {
        if let volumeInfo = item["volumeInfo"] as? [String: AnyObject] {
        let books = data.map { (jsonDictionary) -> Book in
            let title = volumeInfo["title"] as? String ?? ""
            let publishedData = volumeInfo["publishedDate"] as? String ?? ""
            let authors = volumeInfo["authors"] as? [String] ?? [""]

            let description = volumeInfo["description"] as? String ?? ""
            let newBook = Book(title: title, publishedData: publishedData, description: description)
            return newBook
        }
        return books
        }
    }
    return [Book]()
}

这是非常糟糕的方式。你必须在底部返回Book,因为for循环,和 VolumeInfo是下一个词典,所以我真的不知道如何映射它并获得例如作者,因为它是下一个数组..

一个示例JSON对象:

   {  
   "items":[  
      {  
         "volumeInfo":{  
            "title":"The Ancestor's Tale",
            "subtitle":"A Pilgrimage to the Dawn of Life",
            "authors":[  
               "Richard Dawkins",
               "Yan Wong"
            ]
            "publishedDate":"2016-04-28",
            "description":"A fully updated ",
            "imageLinks":{  
               "smallThumbnail":"http://books.google.com/books/content?id=vzbVCQAAQBAJ&printsec=frontcover&img=1&zoom=5&edge=curl&source=gbs_api",
               "thumbnail":"http://books.google.com/books/content?id=vzbVCQAAQBAJ&printsec=frontcover&img=1&zoom=1&edge=curl&source=gbs_api"
            }
}    
]}

所以当你有String:Value数组时,这很简单,但是如果你有字典VolumeInfo中的字典或像作者这样的字符串数组,你应该如何正确映射?

1 个答案:

答案 0 :(得分:1)

我个人发现使用URLSession在swift中解析对象的方法比较笨拙。每当我可以将AlamofireAlamofireObjectMapper结合使用。

这允许您创建一个简单的对象。例如:

class Book: Mappable {
      var title: String?
      var subtitle: String?
      var description: String?

      required init?(map: Map){

      }

      func mapping(map: Map) {
        title <- map["title"]
        subtitle <- map["subtitle"]
        description <- map["description"]
      }
}

当您发出请求时,您可以使用responseObject方法直接解析您的对象并分配正确的类型。

Alamofire.request(URL).responseObject { (response: DataResponse<Book>) in
  let book = response.result.value
  print(book?.title)
}

对于这个例子,我只解析了一本书。但是这个概念也可以很容易地扩展到数组或嵌套的json对象。我个人觉得这比直接使用URLSession的代码更清晰。