【问题标题】:sectioned pickerview chooses all rows in sections with same index分段选择器视图选择具有相同索引的部分中的所有行
【发布时间】:2020-12-30 12:18:02
【问题描述】:

我正在尝试创建一个包含部分的PickerView

每个项目都符合“id”,我用项目的 id 标记Text(这是唯一的,我验证没有冲突)

PickerView 似乎忽略了我分配的标签,并为每个部分选择具有相同对应索引的所有行

我还尝试使用随机 UUID 标记每个项目以检查行为,它似乎继续

struct ExperimentPickerView: View {
    @StateObject var localExperiment = RunningExperiment.active
    @StateObject var experiments = RemoteExperiments.instance
    @State var picked : Int = -1
    var body: some View {
        Picker("active", selection: $picked) {
            ForEach(Array(experiments.experiments.enumerated()), id: \.offset) { i,experiment in
                Section(header: Text("\(experiment.name)")) {
                    ForEach(Array(experiment.variations.enumerated()), id: \.offset) { j,variation in
//                        Text("\(variation.name) \(variation.id)").tag(variation.id)
                        Text("\(variation.name) \(variation.id)").tag(UUID().description)
                        
                    }
                }
            }
        }.id(picked).onReceive([self.picked].publisher.first()) { (value) in
            print(value) // prints the row number and not the id of the element
            
        }
    }
}


struct Experiment : Codable, Equatable, Identifiable {
    var id: Int {
        var hasher = Hasher()
        name.hash(into: &hasher)
        variations.hash(into: &hasher)
        return hasher.finalize()
    }
    let name : String
    let variations: [Variation]
    enum CodingKeys: String, CodingKey {
        case name
        case variations
    }
    
    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)
        let n = try container.decode(String.self, forKey: .name)
        name = n
        let a = try container.decode(AnyCodable.self, forKey: .variations)
        print(a)
        let b = a.value as! [[String:Any]]
        var vars = [Variation]()
        for v in b {
            if let na = v["name"], let nna = na as? String {
                vars.append(Variation(name: nna, experiment: n))
            }
        }
        variations = vars
        
        
    }
}

struct Variation: Codable, Equatable, Identifiable, Hashable {
    var id: Int {
        var hasher = Hasher()
        name.hash(into: &hasher)
        experiment.hash(into: &hasher)
        let hashValue = hasher.finalize()
        return hashValue
    }
    
    let name: String
    var experiment: String
    
    enum CodingKeys: String, CodingKey {
        case name, experiment
    }
}

【问题讨论】:

    标签: swift swiftui pickerview


    【解决方案1】:

    tag 应该与selection 类型相同,因为它用于匹配选取的行。并且id 在这种情况下不应该被修改,因为它会完全重新初始化选择器。

    提供的代码不可测试,因此这里只是一个解决方案演示(您的场景的简化复制)。

    使用 Xcode 12.1 / iOS 14.1 测试。

    struct Pair: Hashable {
        var section = ""
        var row = -1
    }
    
    struct ExperimentPickerView: View {
        var sections = ["A", "B", "C"]
        var rows = Array(0..<5)
    
        @State private var picked = Pair(section: "B", row: 2) // << here !! (initialised demo)
    
        var body: some View {
            Picker("active", selection: $picked) {
                ForEach(sections, id: \.self) { section in
                    Section(header: Text("\(section)").bold()) {
                        ForEach(rows, id: \.self) { i in
                            Text("\(section) \(i)")
                               .tag(Pair(section: section, row: i))   // << match !!
                        }
                    }
                }
            }.onReceive([self.picked].publisher.first()) { (value) in
                print(value)
            }
        }
    }
    

    【讨论】:

    • 它们都是 int 类型且哈希值不变
    • UUID 只是测试它的行为方式。但是使用“variation.id”会产生一个唯一的 Int,它也与“picked”的类型相同
    • 我将它修改为从 \.offset 开始工作,它开始工作了?‍♂️
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