Question

我已经搜索过这个问题，但是我找不到任何使用最新版Xcode和Swift的解决方案。我使用三个数组：

1. baseArray[Meal]: array filled with every meal. Not locally saved.

2. favoritesArray[Favorite]: with names of all favorite meals, locally saved by the user with NSKeyedArchiver.
3. filteredArray[Meal]: baseArray but filtered for searchterm. In code: 

    (filteredArray = baseArray.filter { $0.type == searchtext }}

我使用tableView中的最后一个数组。如果他们想要查看所有餐点，filteredArray与baseArray相同。

我的问题：我怎样才能得到filteredArray所有喜欢的饭菜（Meal.title == Favorite.name）。我如何组合三个阵列？

我在上周尝试了很多选项，但没有一个可行。我希望你能帮帮我!!

Answer 1

这样做你想要的：

struct Meal {
  let title: String
}

struct Favorite {
  let name: String
}

let meal1 = Meal(title: "Soup")
let meal2 = Meal(title: "Stew")
let meal3 = Meal(title: "Pizza")

let favorite1 = Favorite(name: "Stew")

let baseArray = [meal1, meal2, meal3]
let favoritesArray = [favorite1]

let favoriteNames = favoritesArray.map { $0.name }

let filteredArray = baseArray.filter { favoriteNames.contains($0.title) }

Answer 2

如果我正确理解您的问题，这是您的解决方案。

struct Meal {
    let name: String
}

struct Favorite {
    let name: String
}

let baseArray = [Meal(name: "Meal1"), Meal(name: "Meal2"), Meal(name: "Meal3")]

let favoritesArray = [Favorite(name: "Meal1")]

let searchText = "Meal3"

let filteredArray = baseArray.filter { $0.name == searchText }
print(filteredArray)
// [Meal(name: "Meal3")]

let combinedArray = baseArray.reduce(filteredArray) { array, element in
    // First condition check if the current meal (element) in contained in the favorites
    // Second checks if the favorite meal isn't already in the filteder array
    if favoritesArray.contains(where: { $0.name == element.name }) &&
        !filteredArray.contains(where: { $0.name == element.name }) {
        return array + [element]
    }
    return array
}

print(combinedArray)
// [Meal(name: "Meal3"), Meal(name: "Meal1")]

如何在swift中过滤其他数组值的结构数组？

2 个答案: