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SexyJson

Build Status Pod Version Platform Pod License

SexyJson is Swift5+ json parse open source library quickly and easily, perfect supporting class and struct model, support the KVC model, fully oriented protocol architecture, support iOS and MAC OS X

Objective-c version 👉 WHC_Model

update:Support for compatible swift5.+

Note

  • The definition of model must implement SexyJson protocol
  • If you have any enumeration type must be specified in the definition of model data type and implementation SexyJsonEnumType protocol
  • If you want to use swift3.2, please pod SexyJson '~> 0.0.4'

Require

  • iOS 8.0+ / Mac OS X 10.11+ / tvOS 9.0+
  • Xcode 8.0 or later
  • Swift 5.0

Install

  • CocoaPods: pod 'SexyJson'

Model

This is an example of json:

let json = "{\"age\":25,\"enmuStr\":\"Work\",\"url\":\"https:\\/\\/www.baidu.com\",
            \"subArray\":[{\"test3\":\"test3\",\"test2\":\"test2\",\"cls\":{\"age\":10,
            \"name\":\"swift\"},\"test1\":\"test1\"},{\"test3\":\"test3\",\"test2\":\"test2\",
            \"cls\":{\"age\":10,\"name\":\"swift\"},\"test1\":\"test1\"}],\"color\":\"0xffbbaa\",
            \"nestArray\":[[{\"test\":\"test1\"},{\"test\":\"test2\"}],[{\"test\":\"test3\"},
            {\"test\":\"test4\"}]],\"enmuInt\":10,\"sub\":{\"test1\":\"test1\",\"test2\":\"test2\",
            \"test3\":\"test3\"},\"height\":175,\"intArray\":[1,2,3,4],\"name\":\"吴海超\",
            \"learn\":[\"iOS\",\"android\",\"js\",\"nodejs\",\"python\"]}"

The model class:

enum WorkEnum: String,SexyJsonEnumType {
    case null = "nil"
    case one = "Work"
    case two = "Not Work"
}

enum IntEnum: Int,SexyJsonEnumType {
    case zero = 0
    case hao = 10
    case xxx = 20
}

struct Model :SexyJson {

    var age: Int = 0
    var enmuStr: WorkEnum!
    var url: URL!
    var subArray: [SubArray]!
    var color: UIColor!
    var nestArray: [[NestArray]]?
    var enmuInt: IntEnum = .xxx
    var sub: Sub!
    var height: Int = 0
    var intArray: [Int]!
    var name: String!
    var learn: [String]!

    /// Model mapping
    public mutating func sexyMap(_ map: [String : Any]) {
        age      <<<   map["age"]
        enmuStr  <<<   map["enmuStr"]
        url      <<<   map["url"]
        subArray <<<   map["subArray"]
        color    <<<   map["color"]
        nestArray <<<  map["nestArray"]
        enmuInt   <<<  map["enmuInt"]
        sub       <<<  map["sub"]
        height    <<<  map["height"]
        intArray  <<<  map["intArray"]
        name      <<<  map["name"]
        learn     <<<  map["learn"]
    }
}

You don't need to manually create the SexyJson model class you can use open source tools with the help of WHC_DataModel.app automatically created SexyJson model

Usage

Json is converted into a model object(json -> model)

let model = Model.sexy_json(json)

Model object converted into the dictionary(model -> dictionary)

let dictionary = model.sexy_dictionary()

Model object converted into the json string(model -> json)

let jsonStr = model.sexy_json()

SexyJson support json parse the key path

let subArrayModel = SubArray.sexy_json(json, keyPath: "subArray[0]")
let subNestArray = NestArray.sexy_json(json, keyPath: "nestArray[0][0]")
let test = String.sexy_json(json, keyPath: "nestArray[0][0].test")

Json is converted into a model array object(json -> [model])

let arrayModel = [Model].sexy_json(json)

Model object array converted into the array([model] -> array)

let array = arrayModel.sexy_array()

Model object array converted into the json string([model] -> json)

let arrayJson = arrayModel.sexy_json()

SexyJson support model kvc( Model class implement Codable protocol )

let sub = Sub.sexy_json(json, keyPath: "sub")
if let modelCodingData = try? JSONEncoder().encode(modelCoding) {
    if let modelUncoding = try? JSONDecoder().decode(Sub.self, from: modelCodingData) {
        print("modelUncodingJson = \(modelUncoding.sexy_json()!)")
    }
}

Prompt

If you want to view the analytical results, please download this demo to check the specific usage

Licenses

All source code is licensed under the MIT License.