分段选择器视图选择具有相同索引的部分中的所有行

Posted 2023-02-24

【中文标题】分段选择器视图选择具有相同索引的部分中的所有行

【英文标题】：sectioned pickerview chooses all rows in sections with same index

【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
    

【参考方案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?‍♂️