使用Codable时链接领域对象

问题描述 投票:0回答:1

我想将Section链接到我的Category模型。我仅在JSON响应中获取部分ID,所以我使用编码器尝试这样做,但没有用

以下解决方案无效

public required convenience init(from decoder: Decoder) throws {
    self.init()
    let container = try decoder.container(keyedBy: CodingKeys.self)
    self.id = try container.decode(Int.self, forKey: .id)
    self.name = try container.decode(String.self, forKey: .name)
    self.color = try container.decodeIfPresent(String.self, forKey: .color) ?? ""
    let sectionId = try container.decode(Int.self, forKey: .section)
    let section = try! Realm().object(ofType: Section.self, forPrimaryKey: sectionId)
    self.section = section

}

我的解决方案,但我不喜欢它每次都会运行查询的事实

final class Category : Object, Codable {

@objc dynamic var id: Int = 0
@objc dynamic var name: String = ""
@objc dynamic var color: String? = ""
@objc dynamic var sectionId: Int = 0
var section: Section? {
    return self.realm?.object(ofType: Section.self, forPrimaryKey: sectionId)
}

我确信必须有更好的方法。任何线索表示赞赏。

swift realm codable
1个回答
1
投票

如果您对section属性使用惰性变量,则查询将仅运行一次。不利的一面是,如果您观察到类别对象的更改,并且相应的Section对象发生更改,则不会收到通知。

class Category: Object {
    // ...
    @objc dynamic var sectionId: Int = 0

    lazy var section: Section? = {
        return realm?.object(ofType: Section.self, forPrimaryKey: sectionId)
    }()

    override static func ignoredProperties() -> [String] {
        return ["section"]
    }
}
© www.soinside.com 2019 - 2024. All rights reserved.