解决 Swift 中过多的 else-if 语句 3

Question

问题：给定函数的输入，测试每个用户以确保他们符合以下条件： 1. users数组中的每个用户不能与当前用户共享一个聊天室。（聊天室对象有两个属性 'firstUserId' 和 'secondUserId'.
2.users数组中的每个用户都不是当前用户。 3. users 数组中的每个用户都在当前用户的 5 英里半径范围内。

在完成处理程序的调用视线中，我检查 User 对象的值是否为 true，如果是，我将其作为潜在匹配项显示给当前用户。

现在，我很快就用暴力破解了这个解决方案，但每次看到它都会感到畏缩。看起来效率很低。非常感谢有关更优雅解决方案的任何提示！

typealias validUsersCompletionHandler = (_ users: [User: Bool]) -> Void

private func validateNewUsers(currentUser: User, users: [User], chatrooms: [Chatroom], completionHandler: validUsersCompletionHandler?) {

    var results: [User: Bool] = [:]

    let currentUserCoords = CLLocation(latitude: currentUser.latitude, longitude: currentUser.longitude)

    for user in users {
        let newUserCoords = CLLocation(latitude: user.latitude, longitude: user.longitude)
        let distance = currentUserCoords.distance(from: newUserCoords)
        // // 1 mile = 1609 meters, 8046.72 = 5 miles.
        for chatroom in chatrooms {
            if currentUser.id == chatroom.firstUserId && user.id == chatroom.secondUserId {
                results[user] = false
            } else if currentUser.id == chatroom.secondUserId && user.id == chatroom.firstUserId {
                results[user] =  false
            } else if user.id == currentUser.id {
                results[user] = false
            } else if distance > 8046.72 {
                results[user] = false
            } else {
                results[user] = true
            }
        }
    }
    completionHandler?(results)
}

// ********************************************** ******************************

// 下面是我修改后的方法。我想稍微优雅一点？

// ********************************************** ******************************

typealias validUsersCompletionHandler = (_ users: [User: Bool]) -> Void

private func validateNewUsers(currentUser: User, users: [User], chatrooms: [Chatroom], completionHandler: validUsersCompletionHandler?) {

    var results: [User: Bool] = [:]

    var isInRange = false

    var distance: Double = 0 {
        didSet {
            if distance > 8046.72 {
                isInRange = false
            } else {
                isInRange = true
            }
        }
    }

    let currentUserCoords = CLLocation(latitude: currentUser.latitude, longitude: currentUser.longitude)

    let currentUserId = currentUser.id

    for user in users {

        let userId = user.id

        let newUserCoords = CLLocation(latitude: user.latitude, longitude: user.longitude)

        distance = currentUserCoords.distance(from: newUserCoords)
        // // 1 mile = 1609 meters, 8046.72 = 5 miles.

        for chatroom in chatrooms {

            switch (currentUserId, userId, isInRange) {

            case (chatroom.firstUserId,chatroom.secondUserId, _), (_, _, false),(chatroom.secondUserId, chatroom.firstUserId, _), (_, currentUserId, _): results[user] = false

            default: results[user] = true

            }
        }

    }

    completionHandler?(results)

}

}

Answer 1

你可以用 switch 替换 if 语句...或者你可以用 (currentUserId, userId)

 //always check for optionals
    guard let currentUserId = currentUser.id, let userId = user.id, else{
    return
    }
   //The switch should have this format:
    switch (currentUserId, userId){
    //currentUserId == chatroom.firstUserId,  userId = chatroom.secondUserId)
    case (chatroom.firstUserId,chatroom.secondUserId):
    //do your things
    break
    case (chatroom.secondUserId,firstUserId):
    //do other things
    break
    default:
    break
    }

您甚至可以使用带声明的大小写或比较更多选项：

    switch value{
    case let x where value > 10:
    //When value is bigger than 10..etc
    default:
    break
    }

为了更好的使用，见：https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/ControlFlow.html

祝编码愉快:)

Answer 2

或者，您可以将元组值存储在 struc 中的某处（应该稍微更改一下逻辑），但您可以这样做：

struct myBeautifulCases{
static let userIsCurrent = (_,currentUserId,_)
static let sameChatroom = (chatroom.secondUserId, chatroom.firstUserId, _)
static let sameChatroomAlt = (chatroom.secondUserId, chatroom.firstUserId, _)
static let isWithinRange = case (_, _, false)

}

现在你可以用变量名覆盖丑陋的元组，但你应该稍微改变一下逻辑:)祝你好运和成功的项目:)

解决 Swift 中过多的 else-if 语句 3

Resolving too many else-if statements in Swift 3

if-statement

nested-loops

swift