如何使用快速编码根据特定的排序顺序对自定义对象进行排序

Posted 2023-03-14

【中文标题】如何使用快速编码根据特定的排序顺序对自定义对象进行排序

【英文标题】：How to sort custom objects according to a specific sort order using swift codable

【发布时间】：2019-01-07 17:30:23

【问题描述】：

如何根据特定的排序顺序对自定义对象进行快速编码。 我正在使用 swift 可编码协议。

    "data": 
        "team": [
              
                "name": "XYZ",
                "designation": "Managing director - South Asia"
              ,
              
                "name": "XYZ",
                "designation": "Managing director - South Asia"
              
            ],
        "recruiterName": "Lorium Ipsem text",
        "respondsQuickly": "1"
    ,
    "metaData": 
                "sortOrder": [
                  "respondsQuickly",
                  "recruiterName",
                  "team"
                ]
              

这里我需要根据 sortOrder 对象对数据对象进行排序。 现在这是一个模型对象

struct Welcome: Codable 
let data: DataClass?
let metaData: MetaData?

struct JDDataClass: Codable 
   let team: [Team]?
   let recruiterName, respondsQuickly: String?


struct Team: Codable 
   let name, designation: String?

struct JDMetaData: Codable 
let sortOrder: [String]?

我不能将 All Keys 也作为它的模型。 它现在不是 JSON 或字典，因为它是 Codabale 协议 swift 的自定义对象。

期望的输出：

let myArray = [
        "respondsQuickly": "1",
        "recruiterName": "Lorium Ipsem text",
        "team": [
              
                "name": "XYZ",
                "designation": "Managing director - South Asia"
              ,
              
                "name": "XYZ",
                "designation": "Managing director - South Asia"
              
            ]
        
    ]

【问题讨论】：

您的 json 无效，请发布正确的 json 消息。我假设 JDDataClass 也是数组的一部分，否则在 recruiterNamefor instance 上进行排序是没有意义的

请澄清您的问题。你上一个代码例子看不懂，responseDTO是什么data是什么，你为什么要给一个字符串分配一个int？

@JoakimDanielson 完成更改

这没有任何意义。 "data" 键中的值是 Dictionary，但您想将其转换为 Array，这对我的理解没有任何用处。

【参考方案1】：

这是运行的游乐场代码。你可以检查一下。我希望这将有所帮助。 arrResult 持有你的答案。

struct Welcome: Codable 
let data: JDDataClass?
let metaData: JDMetaData?

struct JDDataClass: Codable 
let team: [Team]?
let recruiterName, respondsQuickly: String?


struct Team: Codable 
let name, designation: String?

struct JDMetaData: Codable 
let sortOrder: [String]?


let multiLineString = """

"data": 
"team": [

"name": "XYZ",
"designation": "Managing director - South Asia"
,

"name": "XYZ",
"designation": "Managing director - South Asia"

],
"recruiterName": "Lorium Ipsem text",
"respondsQuickly": "1"
,
"metaData": 
"sortOrder": [
"respondsQuickly",
"recruiterName",
"team"
]


"""
extension Encodable 
subscript(key: String) -> Any? 
    return dictionary[key]

var dictionary: [String: Any] 
    return (try? JSONSerialization.jsonObject(with: JSONEncoder().encode(self))) as? [String: Any] ?? [:]
 
 
 let decoder = JSONDecoder()

 do 
if let data = multiLineString.data(using: String.Encoding.utf8, allowLossyConversion: false)
  let json = try decoder.decode(Welcome.self, from: data)
    var arrResult = [[String : Any]]()
    if let sortOrder = json.metaData?.sortOrder 
       for item in sortOrder 
        if let data = json.data 
            if let obj = data[item] 
                arrResult.append([item: obj])
            
        
    
        print(arrResult)
 
 
  catch 
print(error)
 

【参考方案2】：

按结构键排序没有任何意义。当访问每个元素的成本相同时，为什么要对对象的变量进行排序。

也许你当前的结构是错误的。

  "data": 
    "team": [
      
        "name": "XYZ",
        "designation": "Managing director - South Asia"
      ,
      
        "name": "XYZ",
        "designation": "Managing director - South Asia"
      
    ],
    "recruiterName": "Lorium Ipsem text",
    "respondsQuickly": "1"
  ,
  "metaData": 
    "sortOrder": [
      "respondsQuickly",
      "recruiterName",
      "team"
    ]
  

如果您希望上述输出具有这样的结构，这会更有意义：

  "data": 
    "keys": [
      "team",
      "recruiterName",
      "respondsQuickly"
    ],
    "values": [
      [
        
          "name": "XYZ",
          "designation": "Managing director - South Asia"
        ,
        
          "name": "XYZ",
          "designation": "Managing director - South Asia"
        
      ],
      "Lorium Ipsem text",
      "1"
    ]
  ,
  "metaData": 
    "sortOrder": [
      "respondsQuickly",
      "recruiterName",
      "team"
    ]
  

不过，如果您想使用当前结构实现此输出，您可以使用类似：Reflection 在运行时读取结构的属性并相应地对其进行排序。不用说，这种解决方案很丑陋，应该避免。