如何从字典中获取数据

Question

我有一张狗的名单，需要取一些数据。在一种情况下，我需要在列表中显示名称行，在其他情况下，我需要来自一只狗的全部或部分数据（名称、性别、速度）。我相当肯定我应该使用数组，尽管我是从字典开始的。我计划添加更多参数，并允许用户添加更多的狗，所以我试图找到最可扩展的选项

struct Dog {
    var name: String
    var gender: String
    var speed: Int
    }

struct MyDogs {
    let myDogs = [
        Dog(name: "Saleks", gender: "Male", speed: 50),
        Dog(name: "Balto", gender: "Male", speed: 70),
        Dog(name: "Mila", gender: "Female", speed: 20)
        ]
 }

Answer 1

警告：我没有可用的IDE，可能有一些语法错误。

作为参考，您演示的是而不是多维数组。3d数组是这样的。

let some3DArray =
    [["Hello", "World"],
    ["This", "Is", "An"],
    ["Multidimensional","Array"]]

要访问示例中的值，根据您的要求，您可以这样做。

//To loop through all the dogs in your array. Useful for your "List"
for dog in yourDogs {
    print(" Name: \(dog.name) "
}

// To find a dog based on some property you can do something like this.
let dog = {
    for dog in yourDogs {
        if dog.name == yourSearchValue {
            return dog
        } else {
            //HANDLE NULL VALUE 
            //What do you want to happen if NO dog is found?
        }
        return null
    }
}

// You can use the values from the array by accessing it directly via an index.
// This can be done with whatever conditional you need to specifically reach.
let specificDog = dogs[3]

// Once you have your copy of the specific dog you want to access.
// You can then get the values of that object.
let dogName = specificDog .name
let dogGender = specificDog .gender
let dogSpeed = specificDog .speed

您的用例似乎在正确的轨道上。阵列将非常有用，并提供最大的灵活性，以便以后添加更多的狗。这可以很容易地处理，例如，通过这样做。你可以在这里了解更多。在Swift中将元素添加到数组中

var yourDogArray = [Dogs]()
yourDogArray.append(Dog(name: "xxx", gender: "female", speed: 20))

这是一种常见的用法，它之所以有效，是因为您填充的列表填充在从0到长度的索引上，这意味着如果您选择列表中的第一项，它将与您在arrayCollection.

func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath {
    let name = yourDogArray[indexPath.row].name
    let gender = yourDogArray[indexPath.row].gender
    let speed = yourDogArray[indexPath.row].speed

    //Do whatever else you need to do here with your data. In your case you'd
    //probably segue to the details view controller and present this data.
    //Read up on Segue and Prepare for Segue to pass data between controllers.
}