修改一个可选/非可选字符串数组

//Array of optional Strings
let array : [String?] = ["1",nil,"2","3","4"]

//Separator String
let separator = ","

//flatMap skips the nil values and then joined combines the non nil elements with the separator
let joinedString = array.flatMap{ $0 }.joined(separator: separator)


//Use Compact map in case of **Swift 4**
    let joinedString = array.compactMap{ $0 }.joined(separator: separator

print(joinedString)

这里是flatMap, compactMap跳过数组中的nil值，并附加其他值以给出一个连接字符串。

在Swift 4中

let array:[String] = ["Apple", "Pear ","Orange"]

array.joined(separator: " ")

如果你想转换自定义对象数组为字符串或逗号分隔字符串(csv)，你可以使用

 var stringIds = (self.mylist.map{$0.id ?? 0}).map{String($0)}.joined(separator: ",")

归功于:莫迪 post:将int数组转换为逗号分隔的字符串

如果你的问题是这样的: tobeFormattedString = ["a"， "b"， "c"] 输出= "abc"

String（tobeFormattedString）

let arrayTemp :[String] = ["Mani","Singh","iOS Developer"]
    let stringAfterCombining = arrayTemp.componentsJoinedByString(" ")
   print("Result will be >>>  \(stringAfterCombining)")

结果将>>> Mani Singh iOS开发者

如果你有字符串数组列表，那么转换为Int

let arrayList = list.map { Int($0)!} 
     arrayList.description

它会给你字符串值