XYJSON

An easy way to create parameters of request.

中文介绍

Requirements

• iOS 8.0+
• Swift 5.1+
• Xcode 11.0+

Installation

Installation with CocoaPods:

pod 'XYJSON'

Installation with Carthage:

github "RayJiang16/XYJSON"

Usage

JSONParameters

Struct or Class conform to JSONParameters, you can use requestParameters for free.

struct Employee: JSONParameters {
    var name: String
    var age: Int
}

let employee = Employee(name: "Tom", age: 21)
print(employee.requestParameters)
// {"name":"Tom", "age":21}

JSONProperty

The property name in your model (name in this case), might not same name in request. You can use @JSONProperty to rename that.

struct Employee: JSONParameters {
    @JSONProperty(name: "employee_name")
    var name: String
    var age: Int
}

let employee = Employee(name: "Tom", age: 21)
print(employee.requestParameters)
// {"employee_name":"Tom", "age":21}

There has an other situation is some API might use int property to represent bool in the request.

eg. {"is_vip":0} or {"is_vip":1}

In the code, you should use bool instead of int. You can use convert to resolve the problem.

struct TestConvert: JSONParameters {
    @JSONProperty(convert: convertIntToString)
    var id: Int
    @JSONProperty(name: "is_vip", convert: convertBoolToInt)
    var isVip: Bool
    @JSONProperty(convert: { obj in
        return obj ? "Yes" : "No"
    })
    var custom: Bool
}

let obj = TestConvert(id: 233, isVip: true, custom: false)
print(obj.requestParameters)
// {"id": "233", "is_vip": 1, "custom": "No"}

convert is a closure: (T) -> Any. T is type of property. XYJSON offers some common convert.

• convertBoolToInt
• convertBoolToIntString
• convertIntToString
• convertStringToInt
• convertDoubleToString

JSONIgnore

You can use @JSONIgnore to ignore the property that you don't want it in request parameters.

struct Employee: JSONParameters {
    @JSONProperty(name: "employee_name")
    var name: String
    var age: Int
    @JSONIgnore()
    var other: String = ""
}

let employee = Employee(name: "Tom", age: 21)
print(employee.requestParameters)
// {"employee_name":"Tom", "age":21}

JSONValue

Enum

If you want to use enum in your model, you need to make the enum conform to JSONValue.

enum Sex: Int, JSONValue {
    case girl = 0
    case boy
}
struct Employee: JSONParameters {
    var name: String
    var sex: Sex
}

let employee = Employee(name: "Tom", sex: .boy)
print(employee.requestParameters)
// {"name":"Tom", "sex":1}

Custom model

if you want to use another struct or class in your model, you need to make the model conform to JSONValue and implement var jsonValue: JSONValue.

enum Sex: Int, JSONValue {
    case girl = 0
    case boy
}
struct Employee: JSONParameters {
    var name: String
    var sex: Sex
    var test: Test = Test()
}
struct Test: JSONValue, JSONParameters {
    var t1: Int = 1
    var t2: String = "t2"
    var jsonValue: JSONValue {
        // You can return Int, Double, Array, Dictionary...
        return requestParameters
    }
}

let employee = Employee(name: "Tom", sex: .boy)
print(employee.requestParameters)
// {"name":"Tom", "sex":1, "test":{"t1":1, "t2":"t2"}}

Note

You need to rewrite init function when you use @JSONProperty or @JSONIgnore. The default init function dosen't work, because @propertyWrapper is struct in real.

In the init function, you should set property which is @JSONProperty or @JSONIgnore in the last if you don't set the default value of the property in your model.

struct Employee: JSONParameters {
    @JSONProperty(name: "employee_name")
    var name: String
    var age: Int
    
    init(name: String,
         age: Int) {
        self.age = age
        self.name = name
    }
}

DO NOT use XYJSON and HandyJSON in one model, otherwise HandyJSON dosen't work. Because @propertyWrapper is struct in real.

License

XYJSON is under MIT license. See the LICENSE file for more info.