更新 CollectionView 中的数据模型以过滤单元格

Question

我有 1 个名为 animalCollectionView 的 CollectionView，其中填充了来自 AnimalData 结构的 100 多个 AnimalModel 对象。在 CollectionView 下方，我有 4 个具有不同动物分类的按钮：Fish、Amphibian、Bird、Reptile。我想要这 4 个按钮来切换 CollectionView 中的适当数据。每个 AnimalModel 对象都有一个名为 animalSelected 的属性，应该相应地切换。

在进步方面，我快到了。我只是在遍历所有数据以更改适当对象上的 animalSelected 属性时遇到问题。

这是数据：

struct AnimalModel {
    var animalName:String
    var animalClassification:Animal

    var animalSelected:Bool

    enum Animal:String {
        case bird
        case fish
        case reptile
        case amphibian
    }
}

struct AnimalData {
    static func allAnimal() -> [AnimalModel] {
        return [
            AnimalModel(animalName: "Salmon", animalClassification: .fish, animalSelected: false),
            AnimalModel(animalName: "Goldfish", animalClassification: .fish, animalSelected: false),
            AnimalModel(animalName: "Guppy", animalClassification: .fish, animalSelected: false),
            AnimalModel(animalName: "Shark", animalClassification: .fish, animalSelected: false),
            AnimalModel(animalName: "Tuna", animalClassification: .fish, animalSelected: false),
            AnimalModel(animalName: "Clownfish", animalClassification: .fish, animalSelected: false),
            AnimalModel(animalName: "Swordfish", animalClassification: .fish, animalSelected: false),
            AnimalModel(animalName: "Frog", animalClassification: .amphibian, animalSelected: false),
            AnimalModel(animalName: "Toad", animalClassification: .amphibian, animalSelected: false),
            AnimalModel(animalName: "Newt", animalClassification: .amphibian, animalSelected: false),
            AnimalModel(animalName: "Salamander", animalClassification: .amphibian, animalSelected: false),
            AnimalModel(animalName: "Ostrich", animalClassification: .bird, animalSelected: false),
            AnimalModel(animalName: "Peacock", animalClassification: .bird, animalSelected: false),
            AnimalModel(animalName: "Eagle", animalClassification: .bird, animalSelected: false),
            AnimalModel(animalName: "Duck", animalClassification: .bird, animalSelected: false),
            AnimalModel(animalName: "Chicken", animalClassification: .bird, animalSelected: false),
            AnimalModel(animalName: "Snake", animalClassification: .reptile, animalSelected: false),
            AnimalModel(animalName: "Chameleon", animalClassification: .reptile, animalSelected: false),
            AnimalModel(animalName: "Lizard", animalClassification: .reptile, animalSelected: false),
            AnimalModel(animalName: "Turtle", animalClassification: .reptile, animalSelected: false)
            // + about 80 more... 
        ]
    }
}

这里是风投

class ViewController: UIViewController, UICollectionViewDelegate, UICollectionViewDataSource {

    @IBOutlet weak var animalCollectionView: UICollectionView!

    @IBOutlet weak var birdOutlet: UIButton!
    @IBOutlet weak var fishOutlet: UIButton!
    @IBOutlet weak var reptileOutlet: UIButton!
    @IBOutlet weak var amphibion: UIButton!

    var all = AnimalData.allAnimal()

    override func viewDidLoad() {
        super.viewDidLoad()
        animalCollectionView.delegate = self
        animalCollectionView.dataSource = self
    }

    @IBAction func birdBtnPressed(_ sender: UIButton) {
//          for bird in all {
//              bird.animalSelected = true
//          }
        all[1].animalSelected = true
        animalCollectionView.reloadData()
    }
    @IBAction func fishBtnPressed(_ sender: UIButton) {
    }
    @IBAction func reptileBtnPressed(_ sender: UIButton) {
    }
    @IBAction func amphibionBtnPressed(_ sender: UIButton) {
    }


    func collectionView(_ collectionView: UICollectionView, numberOfItemsInSection section: Int) -> Int {
        return all.count
    }

    func collectionView(_ collectionView: UICollectionView, cellForItemAt indexPath: IndexPath) -> UICollectionViewCell {
        let cell = collectionView.dequeueReusableCell(withReuseIdentifier: "ANIMAL", for: indexPath) as! AnimalCollectionViewCell
        cell.animalNameLabel.text = all[indexPath.item].animalName
        cell.animalClassificationLabel.text = all[indexPath.item].animalClassification.rawValue

        if all[indexPath.item].animalSelected {
            cell.alpha = 1
        } else {
            cell.alpha = 0.2
        }

        return cell
    }

}

birdButtonPressed 是我尝试的地方。我已经注释掉了：

for bird in all {
    bird.animalSelected = true 
}

因为这给出了错误：Cannot assign to property: 'bird' is a 'let' constant

all[1].animalSelected = true 显然不会遍历鸟类，但我这样做只是为了确保第二个单元格 alpha 发生了变化……它确实做到了！ （见下面的截图）

如何遍历数据以将相应单元格上的 animalSelected 更改为 true？

Answer 1

我首先将 Animal 枚举从 AnimalModel 结构中拉出以实现模块化。它还允许使用枚举案例更好地检查是否相等。

要改变你的价值，你必须做两件事。因为你使用的是一个结构，一个值类型而不是一个引用类型，你必须提供一种方法来改变它的内部值。您可以通过添加一个变异函数来实现这一点，如下所示：

struct AnimalModel {
   var animalName: String
   var animalClassification: Animal
   var animalSelected: Bool

   mutating func setAnimalToSelected(value: Bool) {
       animalSelected = value
   }
}

mutating func 允许您更改结构内部的值。

但在此之前，您必须获取它的特定索引并使用 mutating func 更改该特定结构。

for (index, animal) in all.enumerated() {
   if animal.animalClassification == Animal.bird {
       all[index].animalSelected = true
        //if you only have one section
       let indexPath = IndexPath(row: index, section: 0)
       self.collectionView.reloadItems(at: [indexPath])
   }
}