在 Swift 中，当每对元素都是字符串时，如何遍历一个数组以获取 2 个变量？和一个字符串

Posted 2023-02-24

【中文标题】在 Swift 中，当每对元素都是字符串时，如何遍历一个数组以获取 2 个变量？和一个字符串

【英文标题】：In Swift How do I iterate over an array getting 2 variables when each pair of elements is a String? and a String

【发布时间】：2019-04-27 03:21:53

【问题描述】：

我想遍历一个数组 String?，String 重复对，但我无法正确形成“for case let (a,b) in array”。

我想出的最好方法是创建一个 String?, String 的临时结构并创建一个临时结构的数组，然后对其进行迭代，但我想跳过这一步。

下面是基本示例，最后一个 for 循环显示 Xcode 报告的错误。

class Foo 
    var s1: String?
    var s2: String?
    var s3: String?


let foo = Foo()
foo.s1="Test1"
foo.s2=nil
foo.s3="Test3"

let fooArray = [foo.s1, ", ", foo.s2, "; ", foo.s3,"."]
let fooArray1 = [foo.s1,foo.s2, foo.s3]
var text:String = ""
for case let prop? in fooArray1 
    text = text + prop + " / "

print(text)
// The above works but now I want to use a different separator 
//base on the property name

text=""

for case let (prop, sep) in fooArray  // Error <= Expression Type 
// [String?] is ambiguous without more context

    text = text + prop + sep

print(text)

这是我想出来的

struct temp 
    var prop:String?
    var sep:String
    init(_ prop:String?, _ sep:String) 
        self.prop=prop
        self.sep=sep


let ary:[temp] = [ temp(foo.s1,", "), temp(foo.s2,"; "), temp(foo.s3,".") ]
text = ""
for a in ary 
    if let p = a.prop 
        text = text + p + a.sep
    

print (text)

还有其他方法只使用 for 循环

for (a,b) in fooArray 
...

【问题讨论】：

fooArray 不是 (String, String) 的数组。但是let fooArray = [(foo.s1, ", "), (foo.s2, "; "), (foo.s3,".")] 是一个 (String, String) 的数组

【参考方案1】：

正如@RJE 所指出的，fooArray 的推断类型（在您的代码中定义）是 [String?]。

这是使它工作的一种方法：

class Foo 
    var s1: String?
    var s2: String?
    var s3: String?


let foo = Foo()
foo.s1 = "Test1"
foo.s2 = nil
foo.s3 = "Test3"

let fooArray1  = [foo.s1, foo.s2, foo.s3]
let separators = [", ", "; ", "."]

var text = ""

for i in fooArray1.indices 
    if let p = fooArray1[i] 
        text = text + p + separators[i]
    


print (text)  //Test1, Test3.

或者

let zipped = zip(fooArray1, separators)

let text = zipped.map  tuple -> String in
    if case let (x?, y) = tuple 
        return x + y
     else 
        return ""
    
.joined()

print (text)  //Test1,Test3.

或者

let fooArray = [foo.s1, ", ", foo.s2, "; ", foo.s3, "."]
var text = ""
var step = 1
var index = 0

while index < fooArray.count 
    if let str = fooArray[index] 
        step = 1
        text += str
     else 
        step = 2
    
    index += step


print(text) //Test1, Test3.

---

最好这样定义初始化器：

class Foo 
    var s1: String?
    var s2: String?
    var s3: String?

    init(s1: String?, s2: String?, s3: String?) 
        self.s1 = s1
        self.s2 = s2
        self.s3 = s3
    


let foo = Foo(s1: "Test1", s2: nil, s3: "Test3")

---

P.S：所需的输出似乎更适合Foo 类的description 属性。

【参考方案2】：

感谢您的回答，我希望通过这个问题能够更好地了解如何使用 [for] 参数。但是 while 解决方案是我可能会使用以下修改的解决方案

text = ""
var index = 0 
while index < fooArray.count     
    if let prop = fooArray[index] 
        index += 1
        let sep = fooArray[index]!
        index += 1
        text = text + prop + sep
     else 
        index += 2