Question

在Swift中我想创建一个字典数组（具有多个键值对），然后迭代每个元素

以下是可能字典的预期输出。不知道如何声明和初始化它（有点类似于Ruby中的哈希数组）

 dictionary = [{id: 1, name: "Apple", category: "Fruit"}, {id: 2, name: "Bee", category: "Insect"}]

我知道如何用一个键值对创建一个字典数组。例如：

 var airports: [String: String] = ["YYZ": "Toronto Pearson", "DUB": "Dublin"]

Answer 1

声明一个字典数组，使用：

var arrayOfDictionary: [[String : AnyObject]]  = [["id" :1, "name": "Apple", "category" : "Fruit"],["id" :2, "name": "Microsoft", "category" : "Juice"]]

我在你的字典中看到，你将数字与字符串混合在一起，所以最好使用AnyObject而不是String来表示字典中的数据类型。如果在此代码之后，您不必修改此数组的内容，请将其声明为'let'，否则，请使用'var'

更新：在循环中初始化：

//create empty array
var emptyArrayOfDictionary = [[String : AnyObject]]()
for x in 2...3 {  //... mean the loop includes last value => x = 2,3
    //add new dictionary for each loop
    emptyArrayOfDictionary.append(["number" : x , "square" : x*x ])
}
//your new array must contain: [["number": 2, "square": 4], ["number": 3, "square": 9]]

Answer 2

let dic_1: [String: Int] = ["one": 1, "two": 2]
let dic_2: [String: Int] = ["a": 1, "b": 2]
let list_1 = [dic_1, dic_2]

// or in one step:
let list_2: [[String: Int]] = [["one": 1, "two": 2], ["a": 1, "b": 2]]

for d in list_1 { // or list_2
    print(d)
}

导致

[＆＃34;一个＆＃34;：1，＆＃34;两个＆＃34;：2]

[＆＃34; b＆＃34;：2，＆＃34; a＆＃34;：1]

Answer 3

let airports: [[String: String]] = [["YYZ": "Toronto Pearson", "DUB": "Dublin"]]

for airport in airports {
    print(airport["YYZ"])
}

Swift Dictionary多键值对 - 迭代

3 个答案: