【问题标题】:Searching Multiple Struct Arrays With UISearchController Swift 4使用 UISearchController Swift 4 搜索多个结构数组
【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!

标签: ios arrays search swift4 uisearchcontroller


【解决方案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!
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