我正在尝试使用 ObjectMapper 来使用 JSON 响应。到目前为止,我的回复是这样的:
{
"paramsStructure": [
{
"tiles": {
"0": {
"layout": {
"column": 0,
"colSpan": "1",
"rowSpan": "1",
"row": 0
},
"type": "2"
},
"1": {
"layout": {
"column": 1,
"colSpan": "1",
"rowSpan": "1",
"row": 0
},
"type": "2"
}
},
"title": "...",
"rowCount": "4",
"colCount": "2",
"key": "...",
"icon": "..."
}
]
}
到目前为止,我已经为整个 paramsStructure 和 Single Structure 对象的嵌套集合创建了 StructuresObject。现在我想将图 block 映射到嵌套在 Structure 对象中的 TileStructure 对象集合中,如下所示。
class SingleStructure : Mappable{
var columns: Int = 0
var title: String = ""
var key: String = ""
var icon: String = ""
var tilesStructure : [Int: TileStructure]?
required init?(map: Map) {
}
func mapping(map: Map) {
title <- map["title"]
key <- map["key"]
icon <- map["icon"]
columns <- (map["colCount"], TransformOf<Int, String>(
fromJSON: {item in return Int(item!)},
toJSON: {_ in return "$0"}))
//need here parsing of tilesStructure
}
}
我主要需要将此 JSON tiles 字典映射到 [Int: TileStructure],其中键是字典键,TileStructure 是包含“布局”和“类型”属性的可映射对象。
预先感谢您的帮助:)
编辑!!!
我尝试了 denis_lor 方法,但是当我从 RxAlamofire 运行解析数据时,出现以下异常:
keyNotFound(CodingKeys(stringValue: "tiles", intValue: nil), Swift.DecodingError.Context(codingPath: [], debugDescription: "No value associated with key CodingKeys(stringValue: \"tiles\", intValue: nil) (\"tiles\").", underlyingError: nil))
这就是我调用请求的方式
return RxAlamofire.requestData(.get, GlobalSettings.GET_DEVICE_MAIN_STRUCTURE, parameters: parameters, headers: headers)
.debug()
.mapObject(type: ParamsStructure.self)
这就是我的对象映射器:
extension ObservableType {
public func mapObject<T: Codable>(type: T.Type) -> Observable<T> {
return flatMap { data -> Observable<T> in
let responseTuple = data as? (HTTPURLResponse, Data)
guard let jsonData = responseTuple?.1 else {
throw NSError(
domain: "",
code: -1,
userInfo: [NSLocalizedDescriptionKey: "Could not decode object"]
)
}
let decoder = JSONDecoder()
let object = try decoder.decode(T.self, from: jsonData)
return Observable.just(object)
}
}
我认为问题可能出在编码上,这就是造成那些转义“\”的原因,导致 key 不匹配。
最佳答案
这里使用具有动态键的 json 结构的关键是使用 Dictionary
就像我对 [String:Tile]
所做的那样。
您可以尝试使用新的 Swift4's Codable :
import Foundation
public struct ResultParamsStructure: Codable {
public var paramsStructure: [ParamsStructure] = []
enum CodingKeys: String, CodingKey {
case paramsStructure = "paramsStructure"
}
}
public struct ParamsStructure: Codable {
public var tiles: [String:Tile] = [:]
public var title: String = ""
public var rowCount: String = ""
public var colCount: String = ""
public var key: String = ""
public var icon: String = ""
enum CodingKeys: String, CodingKey {
case tiles = "tiles"
case title = "title"
case rowCount = "rowCount"
case colCount = "colCount"
case key = "key"
case icon = "icon"
}
}
public struct Tile: Codable {
public var layout: Layout?
public var type: String = ""
enum CodingKeys: String, CodingKey {
case layout = "layout"
case type = "type"
}
}
public struct Layout: Codable {
public var column: Int = 0
public var colSpan: String = ""
public var rowSpan: String = ""
public var row: Int = 0
enum CodingKeys: String, CodingKey {
case column = "column"
case colSpan = "colSpan"
case rowSpan = "rowSpan"
case row = "row"
}
}
let jsonString = """
{
"paramsStructure": [
{
"tiles": {
"0": {
"layout": {
"column": 0,
"colSpan": "1",
"rowSpan": "1",
"row": 0
},
"type": "2"
},
"1": {
"layout": {
"column": 1,
"colSpan": "1",
"rowSpan": "1",
"row": 0
},
"type": "2"
}
},
"title": "...",
"rowCount": "4",
"colCount": "2",
"key": "...",
"icon": "..."
}
]
}
"""
let json = jsonString.data(using: .utf8)!
let resultParamsStructure = try? JSONDecoder().decode(ResultParamsStructure.self, from: json)
print(resultParamsStructure?.paramsStructure[0].tiles.keys)
print(resultParamsStructure?.paramsStructure[0].tiles["1"]?.layout?.colSpan)
//# Optional(Dictionary.Keys(["0", "1"]))
//# Optional("1")
您可以在这里尝试上面的代码:http://online.swiftplayground.run/
关于swift - ObjectMapper - 将 JSON 字典映射为嵌套对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55128301/