使用Swift 5新奇的可扩展插值:

extension DefaultStringInterpolation {
    mutating func appendInterpolation(pad value: Int, toWidth width: Int, using paddingCharacter: Character = "0") {
        appendInterpolation(String(format: "%\(paddingCharacter)\(width)d", value))
    }
}

let pieCount = 3
print("I ate \(pad: pieCount, toWidth: 3, using: "0") pies")  // => `I ate 003 pies`
print("I ate \(pad: 1205, toWidth: 3, using: "0") pies")  // => `I ate 1205 pies`

Swift 4*及以上，你也可以试试这个:

func leftPadding(valueString: String, toLength: Int, withPad: String = " ") -> String {
        guard toLength > valueString.count else { return valueString }
        
        let padding = String(repeating: withPad, count: toLength - valueString.count)
        return padding + valueString
    }

调用函数:

leftPadding(valueString: "12", toLength: 5, withPad: "0")

输出: “00012”

斯威夫特5

@imanuo answers已经很棒了，但如果你正在使用一个充满数字的应用程序，你可以考虑这样的扩展:

extension String {

    init(withInt int: Int, leadingZeros: Int = 2) {
        self.init(format: "%0\(leadingZeros)d", int)
    }

    func leadingZeros(_ zeros: Int) -> String {
        if let int = Int(self) {
            return String(withInt: int, leadingZeros: zeros)
        }
        print("Warning: \(self) is not an Int")
        return ""
    }
    
}

这样你可以在任何地方打电话:

String(withInt: 3) 
// prints 03

String(withInt: 23, leadingZeros: 4) 
// prints 0023

"42".leadingZeros(2)
// prints 42

"54".leadingZeros(3)
// prints 054

如果您只使用格式字符串处理数字，那么其他答案都很好，但是当您可能有需要填充的字符串时，这个答案也很好(尽管不可否认与所问的问题略有不同，但在精神上似乎是相似的)。此外，如果弦比衬垫长，要小心。

   let str = "a str"
   let padAmount = max(10, str.count)
   String(repeatElement("-", count: padAmount - str.count)) + str

输出“-----a str”

细节

Xcode 9.0.1, swift 4.0

解决方案

Data

import Foundation

let array = [0,1,2,3,4,5,6,7,8]

解决方案1

extension Int {

    func getString(prefix: Int) -> String {
        return "\(prefix)\(self)"
    }

    func getString(prefix: String) -> String {
        return "\(prefix)\(self)"
    }
}

for item in array {
    print(item.getString(prefix: 0))
}

for item in array {
    print(item.getString(prefix: "0x"))
}

解决方案2

for item in array {
    print(String(repeatElement("0", count: 2)) + "\(item)")
}

解决方案3

extension String {

    func repeate(count: Int, string: String? = nil) -> String {

        if count > 1 {
            let repeatedString = string ?? self
            return repeatedString + repeate(count: count-1, string: repeatedString)
        }
        return self
    }
}

for item in array {
    print("0".repeate(count: 3) + "\(item)")
}

下面的代码生成了一个前面填充为0的3位字符串:

import Foundation

var randomInt = Int.random(in: 0..<1000)
var str = String(randomInt)
var paddingZero = String(repeating: "0", count: 3 - str.count)

print(str, str.count, paddingZero + str)

输出:

5 1 005
88 2 088
647 3 647