Ios 快速JSON到多字符串数组的转换

给定以下示例JSON

{
    "filters": [ 
      { "name" : "First Type",
        "types" : ["md", "b", "pb"]},
      { "name" : "Second Type",
        "types" : ["pt", "ft", "t"]},
      { "name" : "Third Type",
        "types" : ["c", "r", "s", "f"]
      }
    ],
    "jobs": [
        { "title":"f",
          "description" : "descrip text",
          "criteria":[ "md", "ft", "s" ],
          "img" : "www1"
        },
        { "title":"boa",
          "description" : "a description",
          "criteria":[ "b", "pb", "f", "ft" ],
          "img" : "www2"
        },
        { "title":"BK",
          "description" : "something here",
          "criteria":[ "md", "pt", "ft", "b", "s" ],
          "img" : "www3"
        }
    ]
 }

（使用Alamofire创建响应） 让responseJSON:JSON=JSON（response.result.value！）

1） 我试图将它们转换为两个字符串数组。一个数组：let filter=[String:[String]]和作业的另一个数组。我该怎么做？（又名给一个人一条鱼）下面是一些示例代码片段，但没有一个能很好地工作

  let filterCategories = responseJSON["filters"].arrayValue.map({
          $0["name"].stringValue
  })

及

---

2） 我如何学习如何正确使用它？（又名教人钓鱼）我一直在阅读文档（），但我很难理解它。我猜最终的答案将使用.map、.stringValue和.arrayValue。最终，尽管我试图避免大量不必要或不可管理的代码

Swift 4提供了现成的JSON解析支持，可以从以下内容开始

根据你可用的结构，我扔到一个操场，并使用

// I was loading the JSON from a file within the Playground's Resource folder
// But basically, you want to end up with a reference to Data
let filePath = Bundle.main.path(forResource:"Source", ofType: "json")
let data = FileManager.default.contents(atPath: filePath!)

struct Filter: Codable {
    let name: String;
    let types: [String];
}

struct Job: Codable {
    let title: String;
    let description: String;
    let criteria: [String];
    let img: String;
}

struct Stuff: Codable {
    let filters: [Filter];
    let jobs: [Job];
}

let decoder = JSONDecoder();
let stuff = try! decoder.decode(Stuff.self, from: data!)

print("Filter:")
for filter in stuff.filters {
    print(filter.name)
    for type in filter.types {
        print(" - \(type)")
    }
}
print("Jobs:")
for job in stuff.jobs {
    print(job.title)
    print(job.description)
    print(job.img)
    for type in job.criteria {
        print(" - \(type)")
    }
}

---

解析结果

Swift 4提供了JSON解析的开箱即用支持，可以从以下内容开始

根据你可用的结构，我扔到一个操场，并使用

// I was loading the JSON from a file within the Playground's Resource folder
// But basically, you want to end up with a reference to Data
let filePath = Bundle.main.path(forResource:"Source", ofType: "json")
let data = FileManager.default.contents(atPath: filePath!)

struct Filter: Codable {
    let name: String;
    let types: [String];
}

struct Job: Codable {
    let title: String;
    let description: String;
    let criteria: [String];
    let img: String;
}

struct Stuff: Codable {
    let filters: [Filter];
    let jobs: [Job];
}

let decoder = JSONDecoder();
let stuff = try! decoder.decode(Stuff.self, from: data!)

print("Filter:")
for filter in stuff.filters {
    print(filter.name)
    for type in filter.types {
        print(" - \(type)")
    }
}
print("Jobs:")
for job in stuff.jobs {
    print(job.title)
    print(job.description)
    print(job.img)
    for type in job.criteria {
        print(" - \(type)")
    }
}

---

要解析结果

可以实现

Codable

协议来解析响应。使用json响应而不是此响应

let url = Bundle.main.url(forResource: "data", withExtension: "json")
let data = NSData(contentsOf: url!)

我用这个做测试

struct Root: Codable {
   let jobs: [Jobs]
   let filters: [Filters]

private enum CodingKeys: String, CodingKey {
    case jobs = "jobs"
    case filters = "filters"
 }
}

struct Filters: Codable {
    let name: String?
    let typees: String?
}

struct Jobs: Codable {
     let title: String?
     let description: String?
     let criteria: [String]?
     let img: String? 
}

let url = Bundle.main.url(forResource: "data", withExtension: "json")
let data = NSData(contentsOf: url!)


do {
      let root = try JSONDecoder().decode(Root.self, from: data as! Data)

      if let name = root.jobs.first?.title {
      print(name)
      }


} catch let error as NSError {

   print(error.description)
}

---

您可以实现

Codable

协议来解析响应。使用json响应而不是此响应

let url = Bundle.main.url(forResource: "data", withExtension: "json")
let data = NSData(contentsOf: url!)

我用这个做测试

struct Root: Codable {
   let jobs: [Jobs]
   let filters: [Filters]

private enum CodingKeys: String, CodingKey {
    case jobs = "jobs"
    case filters = "filters"
 }
}

struct Filters: Codable {
    let name: String?
    let typees: String?
}

struct Jobs: Codable {
     let title: String?
     let description: String?
     let criteria: [String]?
     let img: String? 
}

let url = Bundle.main.url(forResource: "data", withExtension: "json")
let data = NSData(contentsOf: url!)


do {
      let root = try JSONDecoder().decode(Root.self, from: data as! Data)

      if let name = root.jobs.first?.title {
      print(name)
      }


} catch let error as NSError {

   print(error.description)
}

---

如果您使用的是Swift 4，那么请看一看如果您使用的是Swift 4，那么请看一看感觉像伏都教魔法。stuff.filters和stuff.jobs保存着我正在寻找的数据。是的，是的，确实如此，但它（非常）受欢迎地改变了之前的活动中必须做的每一件事，尽管SwiftyJSON非常棒，但这要好得多-我向通过此问题/答案的任何人发出提示：这对我来说一瞬间都不明显，但“结构”JSON的类型需要与所涉及的结构的类型匹配。例如，如果我选择在JSON文件中将“jobs”更改为“abc”，则结构也需要更改（“let jobs”变为“let abc”）。此外，我在构建嵌套数组时使用了这个答案。@Alex You“可能”能够编写自己的“coder”，这在中的“自定义键名”部分中进行了演示；）感觉像巫术。stuff.filters和stuff.jobs保存着我正在寻找的数据。是的，是的，确实如此，但它（非常）受欢迎地改变了之前的活动中必须做的每一件事，尽管SwiftyJSON非常棒，但这要好得多-我向通过此问题/答案的任何人发出提示：这对我来说一瞬间都不明显，但“结构”JSON的类型需要与所涉及的结构的类型匹配。例如，如果我选择在JSON文件中将“jobs”更改为“abc”，则结构也需要更改（“let jobs”变为“let abc”）。此外，我在构建嵌套数组时使用了这个答案。@Alex You“可能”能够编写自己的“coder”，这在中的“自定义键名”部分中进行了演示；）