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?
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
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)")
}
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"))
}
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"
在Xcode 8.3.2, iOS 10.3 这对现在来说很好
Sample1:
let dayMoveRaw = 5
let dayMove = String(format: "%02d", arguments: [dayMoveRaw])
print(dayMove) // 05
Sample2:
let dayMoveRaw = 55
let dayMove = String(format: "%02d", arguments: [dayMoveRaw])
print(dayMove) // 55
假设你想要一个长度为2且前导为0的字段,你可以这样做:
import Foundation
for myInt in 1 ... 3 {
print(String(format: "%02d", myInt))
}
输出:
01 02 03
这需要导入Foundation,所以从技术上讲,它不是Swift语言的一部分,而是Foundation框架提供的功能。请注意,import UIKit和import Cocoa都包含Foundation,所以如果你已经导入了Cocoa或UIKit,就没有必要再次导入它。
格式字符串可以指定多个项的格式。例如,如果你想把3小时15分7秒格式化为03:15:07,你可以这样做:
let hours = 3
let minutes = 15
let seconds = 7
print(String(format: "%02d:%02d:%02d", hours, minutes, seconds))
输出:
03:15:07