我想将Swift中的Int转换为带前导零的字符串。例如，考虑以下代码:
for myInt in 1 ... 3 {
print("\(myInt)")
}
目前的结果是:
1
2
3
但我希望它是:
01
02
03
在Swift标准库中是否有一种干净的方式来做到这一点?
当前回答
对于左填充，添加一个字符串扩展，如下所示:
Swift 5.0 +
extension String {
func padLeft(totalWidth: Int, with byString: String) -> String {
let toPad = totalWidth - self.count
if toPad < 1 {
return self
}
return "".padding(toLength: toPad, withPad: byString, startingAt: 0) + self
}
}
使用这种方法:
for myInt in 1...3 {
print("\(myInt)".padLeft(totalWidth: 2, with: "0"))
}
其他回答
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)")
}
对于左填充，添加一个字符串扩展，如下所示:
Swift 5.0 +
extension String {
func padLeft(totalWidth: Int, with byString: String) -> String {
let toPad = totalWidth - self.count
if toPad < 1 {
return self
}
return "".padding(toLength: toPad, withPad: byString, startingAt: 0) + self
}
}
使用这种方法:
for myInt in 1...3 {
print("\(myInt)".padLeft(totalWidth: 2, with: "0"))
}
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”
Swift 3.0 +
左填充字符串扩展，类似于填充(toLength:withPad:startingAt:)在基础
extension String {
func leftPadding(toLength: Int, withPad: String = " ") -> String {
guard toLength > self.characters.count else { return self }
let padding = String(repeating: withPad, count: toLength - self.characters.count)
return padding + self
}
}
用法:
let s = String(123)
s.leftPadding(toLength: 8, withPad: "0") // "00000123"
