Swift Codable解码手动可选变量

Question

我有以下代码.

import Foundation

let jsonData = """
[
    {"firstname": "Tom", "lastname": "Smith", "age": "28"},
    {"firstname": "Bob", "lastname": "Smith"}
]
""".data(using: .utf8)!

struct Person: Codable {
    let firstName, lastName: String
    let age: String?

    enum CodingKeys : String, CodingKey {
        case firstName = "firstname"
        case lastName = "lastname"
        case age
    }

    init(from decoder: Decoder) throws {
        let values = try decoder.container(keyedBy: CodingKeys.self)
        firstName = try values.decode(String.self, forKey: .firstName)
        lastName = try values.decode(String.self, forKey: .lastName)
        age = try values.decode(String.self, forKey: .age)
    }

}

let decoded = try JSONDecoder().decode([Person].self, from: jsonData)
print(decoded)

问题是它正在崩溃age = try values.decode(String.self, forKey: .age).当我把这个init功能拿出来时它工作正常.错误是No value associated with key age (\"age\")..

关于如何制作可选项的任何想法,如果它不存在则不会崩溃？我还需要init其他功能,但只是举了一个简单的例子来解释发生了什么.

Answer 1

年龄是可选的:

let age: String? 

所以尝试以这种方式解码:

let age: String? = try values.decodeIfPresent(String.self, forKey: .age)