使用 UISearchController Swift 4 搜索多个结构数组

Posted 2023-03-13

【中文标题】使用 UISearchController Swift 4 搜索多个结构数组

【英文标题】：Searching Multiple Struct Arrays With UISearchController Swift 4

【发布时间】：2019-03-05 02:46:00

【问题描述】：

我有两个要搜索的数据数组，但不知道如何使用 UISearchController 进行搜索

struct Fish 
   var name: String
   var price: Int

var fishArray: [Fish] = [   Fish(name: "Bass", price: 1),
                            Fish(name: "Pike", price: 2),
                            Fish(name: "Carp", price: 3),
                            Fish(name: "Bluegill", price: 4),
                            Fish(name: "Catfish", price: 5)]
struct Weapons 
   var name: String
   var price: Int

var weaponsArray: [Weapons] = [   Weapons(name: "Sword", price: 1),
                                  Weapons(name: "Knife", price: 2),
                                  Weapons(name: "Gun", price: 3),
                                  Weapons(name: "Poison", price: 4),
                                  Weapons(name: "Hammer", price: 5)]

搜索发生在这里。仅搜索武器非常有用，但是以后如何添加鱼和其他多个数组。

func filterContentForSearchText(_ searchText: String, scope: String = "All") 
    filteredData = weaponsArray.filter(( weapon : Weapons) -> Bool in
        return weapon.name.lowercased().contains(searchText.lowercased())
    )

    mainTableView.reloadData()

希望这对某人有意义！谢谢！

【问题讨论】：

如果两个结构的属性相似，那你为什么没有一个结构呢？

我实际上不得不问自己同样的问题......但我的例子中同时拥有名称和价格并不是他们唯一的东西......他们每个人都有其他几个独特的属性.话虽如此，我仍然可以让一个结构为我正在做的事情工作。谢谢！

【参考方案1】：

如果它们都有名称和价格，你可以这样做，对于最后一部分，你可以向上转换另一个数组并将它们组合成一个大数组：

protocol Item 
    var name: String  get set 
    var price: Int  get set 


struct Fish: Item 
    var name: String
    var price: Int

var fishArray: [Fish] = [   Fish(name: "Bass", price: 1),
                            Fish(name: "Pike", price: 2),
                            Fish(name: "Carp", price: 3),
                            Fish(name: "Bluegill", price: 4),
                            Fish(name: "Catfish", price: 5)]
struct Weapons: Item 
    var name: String
    var price: Int

var weaponsArray: [Weapons] = [   Weapons(name: "Sword", price: 1),
                                  Weapons(name: "Knife", price: 2),
                                  Weapons(name: "Gun", price: 3),
                                  Weapons(name: "Poison", price: 4),
                                  Weapons(name: "Hammer", price: 5)]

var items: [Item] = (fishArray as [Item]) + (weaponsArray as [Item])
items.filter  (item) -> Bool in
    return item.name.lowercased().contains("C".lowercased())

【讨论】：

这看起来很有希望！我从未使用过协议，但似乎这可能是一个不错的选择！看看这是否有效！谢谢！