使用 UISearchController 时如何删除 sorted/filtered 项

Question

以下代码成功创建，并在 UITableView 中显示 Cars 的列表，稍后您也可以在其中删除购物车。它还提供了一个 UISearchController，您可以在其中成功执行搜索。

我的问题是尝试删除 search/filter 之后的汽车，例如，如果用户搜索位于数组中间的汽车，它将在table 的顶行，但如果 he/she 决定删除它，它将删除 cars 数组中的第一个项目，因为筛选的项目始终位于 [=18] 的顶部=]数组。在这里我没有得到任何错误，但它不会从 cars 数组中删除正确的项目，它总是从 cars 数组中删除第一个项目。

代码如下：

型号

class Car{
    var make = ""
    var model = ""
}

ViewController

class ViewController: UIViewController, UITableViewDataSource, UITableViewDelegate, UISearchResultsUpdating{

    var cars =  Array<Car>()
    var filteredCars =  Array<Car>()
    let searchController = UISearchController(searchResultsController: nil)
    @IBOutlet weak var myTable: UITableView!

    override func viewDidLoad() {
        super.viewDidLoad()
        createCars()

        filteredCars = cars

        searchController.searchResultsUpdater = self
        searchController.obscuresBackgroundDuringPresentation = false
        searchController.searchBar.placeholder = "Search by model"
        navigationItem.searchController = searchController
        definesPresentationContext = true
    }

    func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
        return filteredCars.count
    }

    func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        let cell = tableView.dequeueReusableCell(withIdentifier: "TableCell")! as UITableViewCell
        cell.textLabel?.text = filteredCars[indexPath.row].model
        cell.detailTextLabel?.text = filteredCars[indexPath.row].make
        return cell
    }
    func tableView(_ tableView: UITableView, editActionsForRowAt indexPath: IndexPath) -> [UITableViewRowAction]? {

        let delete = UITableViewRowAction(style: .destructive, title: "Delete") { action, index in
            let alert = UIAlertController(title: "Delete selected car?", message: "This will permanently delete the selected car, do you want to continue?", preferredStyle: UIAlertController.Style.alert)

            alert.addAction(UIAlertAction(title: "Cancel", style: UIAlertAction.Style.cancel, handler: nil))
            alert.addAction(UIAlertAction(title: "Yes", style: UIAlertAction.Style.destructive, handler: { action in
                self.filteredCars.remove(at: indexPath.row)
                self.cars.remove(at: indexPath.row)
                self.myTable.deleteRows(at: [indexPath], with: UITableView.RowAnimation.left)
            }
            ))
            self.present(alert, animated: true, completion: nil)
        }
        return [delete]
    }

    func updateSearchResults(for searchController: UISearchController) {
        if let searchText = searchController.searchBar.text {
            filteredCars = searchText.isEmpty ? cars : cars.filter({(dataString: Car) -> Bool in
                return dataString.model.lowercased().contains(searchText.lowercased())
            })
            myTable.reloadData()
        }
    }
    // create cars manually for demonstration only      
    func createCars(){
        let car1 = Car()
        car1.make = "Ford"
        car1.model = "Explorer"

        //... more cars here

        cars.append(contentsOf: [car1, car2, car3, car4])
    }
}

我尝试了以下方法，但一直收到 Index out of range 错误。

func tableView(_ tableView: UITableView, editActionsForRowAt indexPath: IndexPath) -> [UITableViewRowAction]? {
    let delete = UITableViewRowAction(style: .destructive, title: "Delete") { action, index in
        let alert = UIAlertController(title: "Delete selected car?", message: "This will permanently delete the selected car, do you want to continue?", preferredStyle: UIAlertController.Style.alert)

        alert.addAction(UIAlertAction(title: "Cancel", style: UIAlertAction.Style.cancel, handler: nil))
        alert.addAction(UIAlertAction(title: "Yes", style: UIAlertAction.Style.destructive, handler: { action in

            self.filteredCars.remove(at: indexPath.row)
            self.myTable.deleteRows(at: [indexPath], with: UITableView.RowAnimation.left)

            for i in 0..<self.cars.count {
                if self.cars[i].model == modelToDelete{
                    self.cars.remove(at:i)
                }
            }
        }
        ))
        self.present(alert, animated: true, completion: nil)
    }
    return [delete]
}

搜索后删除项目的正确逻辑是什么？

Answer 1

您必须获取给定汽车在主数组中的索引

alert.addAction(UIAlertAction(title: "Yes", style: UIAlertAction.Style.destructive, handler: { action in
     let carToDelete = self.filteredCars.remove(at: indexPath.row)
     self.cars.remove(at: self.cars.index(of: carToDelete)!)
     tableView.deleteRows(at: [indexPath], with: .left)
}

这需要Car采用Equatable，实现起来还是比较容易的。如果您将 class 更改为 struct，您将免费获得 Equatable。

class Car : Equatable {
    var make = ""
    var model = ""

    static func == (lhs: Car, rhs: Car) -> Bool {
        return lhs.make == rhs.make && lhs.model == rhs.model
    }
}

并且始终使用作为参数传递的 table 视图实例。

使用 UISearchController 时如何删除 sorted/filtered 项

How to delete sorted/filtered items when using UISearchController

ios

swift

uisearchcontroller

型号

ViewController