时间戳

首页 > 解决方案 > 可编码的类与继承

ios - 可编码的类与继承

问题描述

我正在尝试以下代码，但它引发了异常

import Foundation

class SomeBaseClass: Codable {
    let someInt: Int?
    let someString: String?

    init(someInt: Int,someString:String) {
        self.someInt = someInt
        self.someString = someString
    }

    init(someBaseClass: SomeBaseClass) {
        self.someString = someBaseClass.someString
        self.someInt = someBaseClass.someInt
    }
}

class Person: SomeBaseClass {
    let firstName: String?
    let lastName: String?

    init(firstName: String,lastName:String,someBaseClass: SomeBaseClass) {
        self.firstName = firstName
        self.lastName = lastName

        super.init(someBaseClass: someBaseClass)
    }

    enum CodingKeys: String, CodingKey {
        case firstName,lastName
    }

    required init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)

        firstName = try container.decode(String.self, forKey: .firstName)
        lastName = try container.decode(String.self, forKey: .lastName)

        try super.init(from: container.superDecoder())
    }

}

let json = """
{
    "firstName":"John",
    "someInt":2

}
"""

let jsonData = json.data(using: .utf8)!

do {
    let personObject = try JSONDecoder().decode(Person.self, from: jsonData)
    dump(personObject)
} catch let error as NSError {
    print(error)
}

它崩溃并出现以下错误

错误域 = NSCocoaErrorDomain 代码 = 4865 “没有与键 CodingKeys 关联的值（字符串值：“lastName”，intValue：nil）（“lastName”）。” UserInfo={NSDebugDescription=没有与键 CodingKeys 关联的值(stringValue: "lastName", intValue: nil) ("lastName")., NSCodingPath=( )}

我知道lastNameJSON 中缺少它，我已将其标记为可选，但它不像它对结构的工作方式那样工作

有没有一种简单的方法可以做到这一点？

标签： iosswiftcodable

解决方案

你应该使用decodeIfPresent你的可选属性

required init(from decoder: Decoder) throws {
    let container = try decoder.container(keyedBy: CodingKeys.self)
    firstName = try container.decodeIfPresent(String.self, forKey: .firstName)
    lastName = try container.decodeIfPresent(String.self, forKey: .lastName)

    try super.init(from: decoder)
}

还要注意避免另一个错误所需的对 super 的更改。您也可以从

} catch let error as NSError {
    print(error)
}

简单地

} catch {
    print(error)
}

推荐阅读