在Swift中的if语句中使用多个let-as

Question

我正在从字典中解开两个值,在使用它们之前我必须将它们转换为正确的类型.这就是我想出的:

var latitude : AnyObject! = imageDictionary["latitude"]
var longitude : AnyObject! = imageDictionary["longitude"]

if let latitudeDouble = latitude as? Double  {
   if let longitudeDouble = longitude as? Double {
       // do stuff here
   }
}

但是如果让查询合二为一,我想打包两个.所以它会是这样的:

if let latitudeDouble = latitude as? Double, longitudeDouble = longitude as? Double {
    // do stuff here
}

那句法不起作用,所以我想知道是否有一种漂亮的方法可以做到这一点.

Answer 1

Swift 3更新:

以下内容适用于Swift 3:

if let latitudeDouble = latitude as? Double, let longitudeDouble = longitude as? Double {
    // latitudeDouble and longitudeDouble are non-optional in here
}

请务必记住,如果其中一个尝试的可选绑定失败,if-let则不会执行该块内的代码.

注意:这些子句不一定都是'let'子句,你可以用逗号分隔任何一系列布尔检查.

例如:

if let latitudeDouble = latitude as? Double, importantThing == true {
    // latitudeDouble is non-optional in here and importantThing is true
}

Swift 1.2:

Apple可能已经阅读了您的问题,因为您希望的代码在Swift 1.2中正确编译(今天处于测试阶段):

if let latitudeDouble = latitude as? Double, longitudeDouble = longitude as? Double {
    // do stuff here
}

Swift 1.1及更早版本:

这是好消息 - 你可以完全做到这一点.关于两个值的元组的switch语句可以使用模式匹配同时将它们同时转换Double为:

var latitude: Any! = imageDictionary["latitude"]
var longitude: Any! = imageDictionary["longitude"]

switch (latitude, longitude) {
case let (lat as Double, long as Double):
    println("lat: \(lat), long: \(long)")
default:
    println("Couldn't understand latitude or longitude as Double")
}

更新:此版本的代码现在可以正常运行.

Answer 2

使用Swift 3,您可以使用可选的链接,切换语句或可选模式来解决您的问题.

1.使用if let(可选绑定/可选链接)

在斯威夫特编程语言有关可选链接状态:

多个查询可以链接在一起,如果链中的任何链接为零,整个链都会正常失败.

因此,在最简单的情况下,您可以使用以下模式在可选的链接操作中使用多个查询:

let dict = ["latitude": 2.0 as AnyObject?, "longitude": 10.0 as AnyObject?]
let latitude = dict["latitude"]
let longitude = dict["longitude"]

if let latitude = latitude as? Double, let longitude = longitude as? Double {
    print(latitude, longitude)
}

// prints: 2.0 10.0

2.在switch语句中使用元组和值绑定

作为简单的可选链接的替代方法,switch语句在与元组和值绑定一起使用时可以提供细粒度的解决方案:

let dict = ["latitude": 2.0 as AnyObject?, "longitude": 10.0 as AnyObject?]
let latitude = dict["latitude"]
let longitude = dict["longitude"]

switch (latitude, longitude) {
case let (Optional.some(latitude as Double), Optional.some(longitude as Double)):
    print(latitude, longitude)
default:
    break
}

// prints: 2.0 10.0

let dict = ["latitude": 2.0 as AnyObject?, "longitude": 10.0 as AnyObject?]
let latitude = dict["latitude"]
let longitude = dict["longitude"]

switch (latitude, longitude) {
case let (latitude as Double, longitude as Double):
    print(latitude, longitude)
default:
    break
}

// prints: 2.0 10.0

let dict = ["latitude": 2.0 as AnyObject?, "longitude": 10.0 as AnyObject?]
let latitude = dict["latitude"]
let longitude = dict["longitude"]

switch (latitude as? Double, longitude as? Double) {
case let (.some(latitude), .some(longitude)):
    print(latitude, longitude)
default:
    break
}

// prints: 2.0 10.0

let dict = ["latitude": 2.0 as AnyObject?, "longitude": 10.0 as AnyObject?]
let latitude = dict["latitude"]
let longitude = dict["longitude"]

switch (latitude as? Double, longitude as? Double) {
case let (latitude?, longitude?):
    print(latitude, longitude)
default:
    break
}

// prints: 2.0 10.0

3.使用元组if case(可选模式)

if case(可选模式)提供了一种方便的方法来展开可选枚举的值.您可以将它与元组一起使用,以便使用多个查询执行一些可选的链接:

let dict = ["latitude": 2.0 as AnyObject?, "longitude": 10.0 as AnyObject?]
let latitude = dict["latitude"]
let longitude = dict["longitude"]

if case let (.some(latitude as Double), .some(longitude as Double)) = (latitude, longitude) {
    print(latitude, longitude)
}

// prints: 2.0 10.0

let dict = ["latitude": 2.0 as AnyObject?, "longitude": 10.0 as AnyObject?]
let latitude = dict["latitude"]
let longitude = dict["longitude"]

if case let (latitude as Double, longitude as Double) = (latitude, longitude) {
    print(latitude, longitude)
}

// prints: 2.0 10.0

let dict = ["latitude": 2.0 as AnyObject?, "longitude": 10.0 as AnyObject?]
let latitude = dict["latitude"]
let longitude = dict["longitude"]

if case let (.some(latitude), .some(longitude)) = (latitude as? Double, longitude as? Double) {
    print(latitude, longitude)
}

// prints: 2.0 10.0

let dict = ["latitude": 2.0 as AnyObject?, "longitude": 10.0 as AnyObject?]
let latitude = dict["latitude"]
let longitude = dict["longitude"]

if case let (latitude?, longitude?) = (latitude as? Double, longitude as? Double) {
    print(latitude, longitude)
}

// prints: 2.0 10.0

Answer 3

Swift 3.0

if let latitudeDouble = latitude as? Double, let longitudeDouble = longitude as? Double {
    // do stuff here
}