[英]How do I convert this JSON into a Swift structure using the Decodable protocol?
Note : I have already looked at this question -> How do I use custom keys with Swift 4's Decodable protocol? 注意 :我已经看过这个问题-> 如何在Swift 4的Decodable协议中使用自定义键? But it does not explain how to Encode/Decode enums 但是它没有解释如何对枚举进行编码/解码
Here is the structure I want: 这是我想要的结构:
struct MyStruct: Decodable {
let count: PostType
}
enum PostType: Decodable {
case fast(value: Int, value2: Int)
case slow(string: String, string2: String)
}
Now i know how i want my struct to look, the problem is: 现在我知道我希望我的结构看起来如何,问题是:
init
function should look like inside of the PostType
enum. 我不知道init
函数在PostType
枚举内部应该是什么样子。 I use the code below to help me construct the JSON quickly. 我使用下面的代码来帮助我快速构造JSON。
let jsonData = """
{
"count": {
"fast" :
{
"value": 4,
"value2": 5
}
}
}
""".data(using: .utf8)!
// Decoding
do {
let decoder = JSONDecoder()
let response = try decoder.decode(MyStruct.self, from: jsonData)
print(response)
} catch {
print(error)
}
Can anyone help me with this? 谁能帮我这个?
Edit1 my JSON looks like this Edit1我的JSON看起来像这样
{
"count": {
"fast" :
{
"value": 4,
"value2": 5
}
}
}
What should the init
look like in the PostType enum
? init
在PostType enum
应该是什么样子?
Since an enum
with associated types does not match any JSON type you need a bit more handwork and write a custom mapping. 由于具有关联类型的enum
与任何JSON类型都不匹配,因此您需要更多的工作并编写自定义映射。
The following code covers three options. 以下代码涵盖了三个选项。
First of all the enum must not conform to Codable
首先,枚举必须不符合Codable
enum PostType {
case fast(value: Int, value2: Int)
case middle(bool: Bool)
case slow(string: String, string2: String)
}
The case fast
uses an array, middle
a sub-dictionary, slow
separate key / value pairs. fast
使用一个数组, middle
一个子字典, slow
单独的键/值对。
Then declare the MyStruct
struct, adopt Codable
and declare the type
然后声明MyStruct
结构,采用Codable
并声明type
struct MyStruct : Codable {
var type : PostType
This solution requires custom keys 此解决方案需要自定义键
enum CodingKeys: String, CodingKey {
case value, string, string2, middle
}
The encode
method switch
es on the cases and creates the appropriate types encode
方法switch
案例并创建适当的类型
func encode(to encoder: Encoder) throws {
var container = encoder.container(keyedBy: CodingKeys.self)
switch type {
case .fast(let value, let value2) :
try container.encode([value, value2], forKey: .value)
case .slow(let string, let string2) :
try container.encode(string, forKey: .string)
try container.encode(string2, forKey: .string2)
case .middle(let bool):
try container.encode(["bool" : bool], forKey: .middle)
}
}
In the decode
method you can distinguish the cases by the passed keys, make sure that they are unique. 在decode
方法中,您可以通过传递的键来区分大小写,并确保它们是唯一的。
init(from decoder: Decoder) throws {
let values = try decoder.container(keyedBy: CodingKeys.self)
let allKeys = values.allKeys
if allKeys.contains(.middle) {
let value = try values.decode([String:Bool].self, forKey: .middle)
type = PostType.middle(bool: value["bool"]!)
} else if allKeys.contains(.value) {
let value = try values.decode([Int].self, forKey: .value)
type = PostType.fast(value: value[0], value2: value[1])
} else {
let string = try values.decode(String.self, forKey: .string)
let string2 = try values.decode(String.self, forKey: .string2)
type = PostType.slow(string: string, string2: string2)
}
}
}
Although some keys are hard-coded the first option seems to be the most suitable one. 尽管有些键是硬编码的,但第一个选项似乎是最合适的。
Finally an example to use it: 最后是一个使用它的示例:
let jsonString = "[{\"value\": [2, 6]}, {\"string\" : \"foo\", \"string2\" : \"bar\"}, {\"middle\" : {\"bool\" : true}}]"
let jsonData = jsonString.data(using: .utf8)!
do {
let decoded = try JSONDecoder().decode([MyStruct].self, from: jsonData)
print("decoded:", decoded)
let newEncoded = try JSONEncoder().encode(decoded)
print("re-encoded:", String(data: newEncoded, encoding: .utf8)!)
} catch {
print(error)
}
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