在索引路径 SWIFT 处获取枚举值

Question

我有这个模型，我在一个 tableview 上使用并填充它：

struct TestSection {
   var title: String
   var options: [TestOption]
}

struct TestOption {
   var field: String
   var placeholder: String
}

var models = [TestSection]()

override func viewDidLoad() {
        super.viewDidLoad()

models.append(TestSection(title: "Test", options: [
            TestOption(placeholder: "Test"),
            TestOption(placeholder: "Test 2"),
            TestOption(placeholder: "Test 3")
        ]))
}

使用委托传回一些数据后，我可以更改特定索引处的模型值，如下所示： 模型[0].options[1].placeholder = value

我想用枚举实现同样的目标。我的带有枚举的模型如下所示：

struct TestSection {
   var title: String
   var options: [TestOptionType]
}

enum TestOptionType {
   case testOneCell(model: TestOptionOne)
   case testTwoCell(model: TestOptionTwo)
   case testThreeCell(model: TestOptionThree)
}

struct TestOptionOne {
   var field: String
   var placeholder: String
}

struct TestOptionTwo {
   var tag: Int
   var placeholder: String
   var handler: (() -> Void)
}

struct TestOptionThree {
   var firstName: String
   var lastName: String
   var handler: (() -> Void)
}

models.append(TestSection(title: "Description", options: [
            .testOneCell(model: TestOptionOne(field: "Test", placeholder: "Test")),
            .testTwoCell(model: TestOptionTwo(tag: 1, placeholder: "Test", handler: {
             ///handle handler
             })),
            .testThreeCell(model: TestOptionThree(firstName: "Test", lastName: "Test", handler: {
             ///handle handler
             }))
        ]))

如何使用 models[0].options[0] 更改我的结构之一的值。当我尝试它时，我只能将 'self' 添加到诸如 'models[0].options[0].self' 这样的选项的末尾，但这不是我想要的。

Answer 1

由于TestOption 是一个枚举，models[0].options[0] 的值可能是三种情况之一，并且在每种情况下，都有一个不同的结构值与之关联。

您似乎想要设置特定选项类型之一的属性（TestOptionOne、TestOptionTwo 或 TestOptionThree）。好吧，如果不确保 存在 实际上是与枚举值相关联的那种类​​型的值，您就无法做到这一点。例如，要设置models[0].options[0].field，您必须首先确保models[0].options[0] 是testOneCell 的情况。为此：

if case .testOneCell(var optionOne) = models[0].options[0] {

    // after we do the check, we can set the entire enum, but only changing field to a new value
    models[0].options[0] = .testOneCell(TestOptionOne(
        field: "This is a new value for field", placeholder: optionOne.placeholder
    ))
} else {
    // You should decide what to do when the enum is not of case testOneCell
}

每次都这样做很烦人，所以你可以把它提取到一个变异函数中并放入枚举中：

mutating func setOptionOne(mutator: (inout TestOptionOne) -> Void) {
    if case .testOneCell(var optionOne) = self {
        mutator(&optionOne)
        self = .testOneCell(model: optionOne)
    } else {
        // You should decide what to do when this is called and the enum wasn't of case testOneCell
    }
}

mutating func setOptionTwo(mutator: (inout TestOptionTwo) -> Void) {
    if case .testTwoCell(var optionTwo) = self {
        mutator(&optionTwo)
        self = .testTwoCell(model: optionTwo)
    } else {
        // You should decide what to do when this is called and the enum wasn't of case testTwoCell
    }
}

mutating func setOptionThree(mutator: (inout TestOptionThree) -> Void) {
    if case .testThreeCell(var optionThree) = self {
        mutator(&optionThree)
        self = .testThreeCell(model: optionThree)
    } else {
        // You should decide what to do when this is called and the enum wasn't of case testThreeCell
    }
}

用法：

models[0].options[0].setOptionOne { $0.field = "This is a new value for field" }