Ios 如何使用动态类型序列化JSON
我需要序列化以下类型的JSON响应。哪个有不同的类型Ios 如何使用动态类型序列化JSON,ios,json,swift,object,serialization,Ios,Json,Swift,Object,Serialization,我需要序列化以下类型的JSON响应。哪个有不同的类型 [{ "type": "respiration_rate", "value": 45 } { "type": "blood_pressure", "value": { hg: 50 ,mm:120 } }] 我用于序列化上层json的类是 class Template: Codable { var type: String? var value: Double? private enum CodingKeys: St
[{
"type": "respiration_rate",
"value": 45
}
{ "type": "blood_pressure",
"value": { hg: 50 ,mm:120 }
}]
我用于序列化上层json的类是
class Template: Codable {
var type: String?
var value: Double?
private enum CodingKeys: String, CodingKey {
case type
case value
}
}
如何序列化
值
它是双精度还是动态对象 以下是您需要的代码:
class Template: Codable {
let type: String?
let value: Value?
private enum CodingKeys: String, CodingKey {
case type
case value
}
typealias ValueDictionary = Dictionary<String, Int>
enum Value: Codable {
case double(Double)
case object(ValueDictionary)
init(from decoder: Decoder) throws {
let container = try decoder.singleValueContainer()
if let x = try? container.decode(Double.self) {
self = .double(x)
return
}
if let x = try? container.decode(ValueDictionary.self) {
self = .object(x)
return
}
throw DecodingError.typeMismatch(Value.self, DecodingError.Context(codingPath: decoder.codingPath, debugDescription: "Wrong type for ValueUnion"))
}
func encode(to encoder: Encoder) throws {
var container = encoder.singleValueContainer()
switch self {
case .double(let x):
try container.encode(x)
case .object(let x):
try container.encode(x)
}
}
}
}
类模板:可编码{
让类型:字符串?
让价值:价值?
私有枚举编码键:字符串,编码键{
案例类型
案例价值
}
typealias ValueDictionary=字典
枚举值:可编码{
双箱(双箱)
案例对象(ValueDictionary)
init(来自解码器:解码器)抛出{
let container=尝试解码器。singleValueContainer()
如果让x=try?container.decode(Double.self){
self=.double(x)
返回
}
如果让x=try?container.decode(ValueDictionary.self){
self=.object(x)
返回
}
抛出DecodingError.typeMismatch(Value.self,DecodingError.Context(codingPath:decoder.codingPath,debugDescription:“ValueUnion的类型错误”))
}
func encode(到编码器:编码器)抛出{
var container=encoder.singleValueContainer()
切换自身{
双格(x):
尝试容器编码(x)
case.object(设x):
尝试容器编码(x)
}
}
}
}
您可以使用带有关联值的枚举,而不是类,因为JSON格式似乎是异构的
enum Value: Decodable {
case respirationRate(Double)
case bloodPressure(hg: Double, mm: Double)
private struct BloodPresure: Decodable {
let hg: Double
let mm: Double
}
private enum CodingKeys: String, CodingKey {
case type
case value
}
init(from decoder: Decoder) throws {
let container = try decoder.container(keyedBy: CodingKeys.self)
switch try container.decode(String.self, forKey: .type) {
case "respiration_rate":
self = try .respirationRate(container.decode(Double.self, forKey: .value))
case "blood_pressure":
let details = try container.decode(BloodPresure.self, forKey: .value)
self = .bloodPressure(hg: details.hg, mm: details.mm)
case let type: throw DecodingError.dataCorruptedError(forKey: .value, in: container, debugDescription: "Invalid type: \(type)")
}
}
}
解码json现在很容易:
let json = """
[{
"type": "respiration_rate",
"value": 45
},
{ "type": "blood_pressure",
"value": { "hg": 50 ,"mm":120 }
}]
"""
do {
let values = try JSONDecoder().decode([Value].self, from: json.data(using: .utf8)!)
print(values)
} catch {
print(error)
}
使用
Any?
代替Double?
并在稍后阶段转换为受尊重的类型(使用该值时)使用Any?
抛出不符合协议
编译时错误。我可能会建议您退回到手动解析,以便可以创建适当类型结构的实例。如果您只有这两个选项,呼吸率
将有一个对象的值双
和血压
?试试看