答案 0 :(得分:8)
这只是将来自WWDC 2015的新系统字体介绍视频中的Apple示例代码放入游乐场并使用UILabel使用字体功能呈现纯文本分数。 [更新为Swift 4]
//: Playground - noun: a place where people can play
import UIKit
import CoreGraphics
let pointSize : CGFloat = 60.0
let systemFontDesc = UIFont.systemFont(ofSize: pointSize,
weight: UIFont.Weight.light).fontDescriptor
let fractionFontDesc = systemFontDesc.addingAttributes(
[
UIFontDescriptor.AttributeName.featureSettings: [
[
UIFontDescriptor.FeatureKey.featureIdentifier: kFractionsType,
UIFontDescriptor.FeatureKey.typeIdentifier: kDiagonalFractionsSelector,
],
]
] )
let label = UILabel(frame: CGRect(x: 0, y: 0, width: 200, height: 100))
label.font = UIFont(descriptor: fractionFontDesc, size:pointSize)
label.text = "12/48" // note just plain numbers and a regular slash
只需轻拍操场上的眼睛,您就会看到一个美丽的部分。
答案 1 :(得分:4)
我必须在App中做类似的事情。我在公共分数和相关的unicode字符之间创建了一个映射,如下所示:
enum Fraction: Double {
case Eighth = 0.125
case Quarter = 0.25
case Third = 0.333333333333333
case Half = 0.5
case TwoThirds = 0.666666666666667
case ThreeQuarters = 0.75
}
func localizedStringFromFraction(fraction: Fraction) -> String {
switch fraction {
case .Eighth:
return NSLocalizedString("\u{215B}", comment: "Fraction - 1/8")
case .Quarter:
return NSLocalizedString("\u{00BC}", comment: "Fraction - 1/4")
case .Third:
return NSLocalizedString("\u{2153}", comment: "Fraction - 1/3")
case .Half:
return NSLocalizedString("\u{00BD}", comment: "Fraction - 1/2")
case .TwoThirds:
return NSLocalizedString("\u{2154}", comment: "Fraction - 2/3")
case .ThreeQuarters:
return NSLocalizedString("\u{00BE}", comment: "Fraction - 3/4")
}
}
如果您需要支持更多分数,可以找到映射here。
答案 2 :(得分:0)
以下是任意分数的Swift解决方案:
enum ScriptType {
case Superscript
case Subscript
}
func createSuperOrSubscriptDigit(character:Character, type:ScriptType) -> Character {
switch character {
case "0": return type == .Superscript ? "\u{2070}" : "\u{2080}"
case "1": return type == .Superscript ? "\u{00b9}" : "\u{2081}"
case "2": return type == .Superscript ? "\u{00b2}" : "\u{2082}"
case "3": return type == .Superscript ? "\u{00b3}" : "\u{2083}"
case "4": return type == .Superscript ? "\u{2074}" : "\u{2084}"
case "5": return type == .Superscript ? "\u{2075}" : "\u{2085}"
case "6": return type == .Superscript ? "\u{2076}" : "\u{2086}"
case "7": return type == .Superscript ? "\u{2077}" : "\u{2087}"
case "8": return type == .Superscript ? "\u{2078}" : "\u{2088}"
case "9": return type == .Superscript ? "\u{2079}" : "\u{2089}"
default: return character
}
}
func createSuperOrSubscriptDigits(string:String, type:ScriptType) -> String {
return String(string.characters.map() { createSuperOrSubscriptDigit($0, type: type) })
}
extension String {
func createSuperscriptDigits() -> String {
return createSuperOrSubscriptDigits(self, type: .Superscript)
}
func createSubscriptDigits() -> String {
return createSuperOrSubscriptDigits(self, type: .Subscript)
}
}
func fractionString(numerator:String, denominator:String) -> String {
return numerator.createSuperscriptDigits() + "\u{2044}" + denominator.createSubscriptDigits()
}
答案 3 :(得分:0)
@ KennethBruno的答案有效,但我发现它有点不雅。这是我的版本:
let superscriptDigits = Array("⁰¹²³⁴⁵⁶⁷⁸⁹".characters)
let subscriptDigits = Array("₀₁₂₃₄₅₆₇₈₉".characters)
func vulgarFractionWithNumerator(numerator: UInt, denominator: UInt) -> String {
let zeroBias = UnicodeScalar("0").value
let supers = "\(numerator)".unicodeScalars.map { superscriptDigits[Int($0.value - zeroBias)] }
let subs = "\(denominator)".unicodeScalars.map { subscriptDigits[Int($0.value - zeroBias)] }
return String(supers + [ "⁄" ] + subs)
}
vulgarFractionWithNumerator(123, denominator: 45678)
结果:
“¹²³/₄₅₆₇₈”