如何将Swift数组转换为字符串？

我知道如何以编程方式做到这一点，但是我敢肯定有一种内置的方式...

我使用的每种语言都有一组对象的默认文本表示形式，当您尝试将Array与字符串连接起来或将其传递给print（）函数时，它会吐出。Apple的Swift语言是否可以有一种轻松地将数组转换为字符串的内置方法，还是在对数组进行字符串化时始终必须明确？

如果数组包含字符串，则可以使用String的join方法：

var array = ["1", "2", "3"]

let stringRepresentation = "-".join(array) // "1-2-3"

在Swift 2中：

var array = ["1", "2", "3"]

let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"

如果要使用特定的分隔符（连字符，空格，逗号等），这可能很有用。

否则，您可以简单地使用description属性，该属性返回数组的字符串表示形式：

let stringRepresentation = [1, 2, 3].description // "[1, 2, 3]"

提示：任何实现Printable协议的对象都有一个description属性。如果您在自己的类/结构中采用该协议，则也使它们易于打印

在Swift 3中

• join成为joined，例如[nil, "1", "2"].flatMap({$0}).joined()
• joinWithSeparator变为joined(separator:)（仅适用于字符串数组）

在Swift 4中

var array = ["1", "2", "3"]
array.joined(separator:"-")

使用Swift 5，您可以根据需要选择以下Playground示例代码之一来解决您的问题。

将Characters 数组转换为String不带分隔符的：

let characterArray: [Character] = ["J", "o", "h", "n"]
let string = String(characterArray)

print(string)
// prints "John"

将Strings 数组转换为String不带分隔符的：

let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: "")

print(string) // prints: "BobDanBryan"

将Strings 数组变成a String，并在单词之间使用分隔符：

let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: " ")

print(string) // prints: "Bob Dan Bryan"

将Strings 数组转换为String，并在字符之间使用分隔符：

let stringArray = ["car", "bike", "boat"]
let characterArray = stringArray.flatMap { $0 }
let stringArray2 = characterArray.map { String($0) }
let string = stringArray2.joined(separator: ", ")

print(string) // prints: "c, a, r, b, i, k, e, b, o, a, t"

将Floats 数组转换为a String，并在数字之间使用分隔符：

let floatArray = [12, 14.6, 35]
let stringArray = floatArray.map { String($0) }
let string = stringArray.joined(separator: "-")

print(string)
// prints "12.0-14.6-35.0"

Swift 2.0 Xcode 7.0 beta 6及更高版本使用joinWithSeparator()代替join()：

var array = ["1", "2", "3"]
let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"

joinWithSeparator 被定义为 SequenceType

extension SequenceType where Generator.Element == String {
    /// Interpose the `separator` between elements of `self`, then concatenate
    /// the result.  For example:
    ///
    ///     ["foo", "bar", "baz"].joinWithSeparator("-|-") // "foo-|-bar-|-baz"
    @warn_unused_result
    public func joinWithSeparator(separator: String) -> String
}

迅捷3

["I Love","Swift"].joined(separator:" ") // previously joinWithSeparator(" ")

在Swift 4中

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

array.joined(separator: " ")

由于没有人提到reduce，因此这里是：

[0, 1, 1, 0].map {"\($0)"}.reduce("") {$0 + $1 } // "0110"

本着函数式编程的精神🤖

更改可选/非可选字符串的数组

//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)

在这里，flatMap，compactMap跳过数组中的nil值，并附加其他值以提供连接的字符串。

我的工作与组件JoinedByString NSMutableArray

var array = ["1", "2", "3"]
let stringRepresentation = array.componentsJoinedByString("-") // "1-2-3"

在Swift 2.2中，您可能必须将数组转换为NSArray才能使用componentsJoinedByString（“，”）

let stringWithCommas = (yourArray as NSArray).componentsJoinedByString(",")

如果要抛空数组中的空字符串。

["Jet", "Fire"].filter { !$0.isEmpty }.joined(separator: "-")

如果还要过滤nil值：

["Jet", nil, "", "Fire"].flatMap { $0 }.filter { !$0.isEmpty }.joined(separator: "-")

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

结果将是>>> Mani Singh iOS Developer

与您描述的Swift等效的是字符串插值。如果您正在考虑像JavaScript这样的事情"x" + array，那么Swift中的等效项是"x\(array)"。

作为一般说明，字符串插值与Printable协议之间存在重要区别。仅某些类符合Printable。每个类都可以以某种方式插入字符串。这在编写泛型函数时很有用。您不必局限于Printable上课。

您可以使用打印功能打印任何对象

或用于\(name)将任何对象转换为字符串。

例：

let array = [1,2,3,4]

print(array) // prints "[1,2,3,4]"

let string = "\(array)" // string == "[1,2,3,4]"
print(string) // prints "[1,2,3,4]"

为创建扩展名Array：

extension Array {

    var string: String? {

        do {

            let data = try JSONSerialization.data(withJSONObject: self, options: [.prettyPrinted])

            return String(data: data, encoding: .utf8)

        } catch {

            return nil
        }
    }
}

对于希伯来语或日语等某些语言，分隔符可能不是一个好主意。尝试这个：

// Array of Strings
let array: [String] = ["red", "green", "blue"]
let arrayAsString: String = array.description
let stringAsData = arrayAsString.data(using: String.Encoding.utf16)
let arrayBack: [String] = try! JSONDecoder().decode([String].self, from: stringAsData!)

对于其他数据类型分别：

// Set of Doubles
let set: Set<Double> = [1, 2.0, 3]
let setAsString: String = set.description
let setStringAsData = setAsString.data(using: String.Encoding.utf16)
let setBack: Set<Double> = try! JSONDecoder().decode(Set<Double>.self, from: setStringAsData!)

如果您有字符串数组列表，则转换为Int

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

它会给你字符串值

对于任何元素类型

extension Array {

    func joined(glue:()->Element)->[Element]{
        var result:[Element] = [];
        result.reserveCapacity(count * 2);
        let last = count - 1;
        for (ix,item) in enumerated() {
            result.append(item);
            guard ix < last else{ continue }
            result.append(glue());
        }
        return result;
    }
}

尝试这个：

let categories = dictData?.value(forKeyPath: "listing_subcategories_id") as! NSMutableArray
                        let tempArray = NSMutableArray()
                        for dc in categories
                        {
                            let dictD = dc as? NSMutableDictionary
                            tempArray.add(dictD?.object(forKey: "subcategories_name") as! String)
                        }
                        let joinedString = tempArray.componentsJoined(by: ",")

对于SWIFT 3：

func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
    if textField == phoneField
    {
        let newString = NSString(string: textField.text!).replacingCharacters(in: range, with: string)
        let components = newString.components(separatedBy: NSCharacterSet.decimalDigits.inverted)

        let decimalString = NSString(string: components.joined(separator: ""))
        let length = decimalString.length
        let hasLeadingOne = length > 0 && decimalString.character(at: 0) == (1 as unichar)

        if length == 0 || (length > 10 && !hasLeadingOne) || length > 11
        {
            let newLength = NSString(string: textField.text!).length + (string as NSString).length - range.length as Int

            return (newLength > 10) ? false : true
        }
        var index = 0 as Int
        let formattedString = NSMutableString()

        if hasLeadingOne
        {
            formattedString.append("1 ")
            index += 1
        }
        if (length - index) > 3
        {
            let areaCode = decimalString.substring(with: NSMakeRange(index, 3))
            formattedString.appendFormat("(%@)", areaCode)
            index += 3
        }
        if length - index > 3
        {
            let prefix = decimalString.substring(with: NSMakeRange(index, 3))
            formattedString.appendFormat("%@-", prefix)
            index += 3
        }

        let remainder = decimalString.substring(from: index)
        formattedString.append(remainder)
        textField.text = formattedString as String
        return false
    }
    else
    {
        return true
    }
}

如果您的问题是这样的：tobeFormattedString = [“ a”，“ b”，“ c”]输出=“ abc”

String(tobeFormattedString)