我具有以下代码结构。如果我省略了codingkey部分,则代码正在运行。我实现了StringConverter来将字符串转换为Int(来自bySundell Swift Side)
struct Video: Codable {
var title: String
var description: String
var url: URL
var thumbnailImageURL: URL
var numberOfLikes: Int {
get { return likes.value }
}
private var likes: StringBacked<Int>
enum CodingKeys: String, CodingKey{
case title = "xxx"
case description = "jjjj"
case url = "url"
case thumbnailImageURL = "jjjjjjjj"
case numberofLikes = "jjjjjkkkk"
}
}
// here the code for converting the likes
protocol StringRepresentable: CustomStringConvertible {
init?(_ string: String)
}
extension Int: StringRepresentable {}
struct StringBacked<Value: StringRepresentable>: Codable {
var value: Value
init(from decoder: Decoder) throws {
let container = try decoder.singleValueContainer()
let string = try container.decode(String.self)
let stringToConvert = string.split(separator: "/").last!.description
guard let value = Value(stringToConvert) else {
throw DecodingError.dataCorruptedError(
in: container,
debugDescription: """
Failed to convert an instance of \(Value.self) from "\(string)"
"""
)
}
self.value = value
}
func encode(to encoder: Encoder) throws {
var container = encoder.singleValueContainer()
try container.encode(value.description)
}}
正如我所说,如果我省略了Codingkeys部分,它不会显示任何错误。我将简单地创建一个结构,从Rest API中获取一个字符串,并希望使用Codable在数据库中转换为Int。 谢谢 阿诺德
答案 0 :(得分:1)
定义CodingKeys
时,需要为每个非可选/未初始化的属性提供密钥,以便编译器知道如何在解码时进行初始化。将此应用到Video
上,它将看起来像这样,
struct Video: Codable {
var title: String
var description: String
var url: URL
var thumbnailImageURL: URL
var numberOfLikes: Int {
return likes.value
}
private var likes: StringBacked<Int>
enum CodingKeys: String, CodingKey{
case title = "xxx"
case description = "jjjj"
case url = "url"
case thumbnailImageURL = "jjjjjjjj"
case likes = "jjjjjkkkk"
}
}
如果仔细观察,枚举中未提供此属性private var likes: StringBacked<Int>
,则CodingKey
会因此引起编译器的抱怨。在这种情况下,我用case likes = "jjjjjkkkk"
更新了枚举,并删除了case numberofLikes = "jjjjjkkkk"
,因为numberofLikes
是一个只读的计算属性,不需要任何解析。