ios 如何在 Swift 中使用下标和上标？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/29225779/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How do I use subscript and superscript in Swift?
提问by Adam Young
I want my UILabelto display text in following manner 6.022*1023. What functions does Swift have for subscript and superscript?
我希望我UILabel以以下方式显示文本 6.022*10 23。Swift 对下标和上标有哪些功能？
回答by Forrest Porter
Most of the answers+examples are in ObjC, but this is how to do it in Swift.
大多数答案+示例都在 ObjC 中，但这是在 Swift 中的方法。
let font:UIFont? = UIFont(name: "Helvetica", size:20)
let fontSuper:UIFont? = UIFont(name: "Helvetica", size:10)
let attString:NSMutableAttributedString = NSMutableAttributedString(string: "6.022*1023", attributes: [.font:font!])
attString.setAttributes([.font:fontSuper!,.baselineOffset:10], range: NSRange(location:8,length:2))
labelVarName.attributedText = attString
This gives me:
这给了我：
In a more detailed explanation:
在更详细的解释中：
- Get
UIFontyou want for both the default and superscript style, superscript must be smaller. - Create a
NSMutableAttributedStringwith the full string and default font. - Add an attribute to the characters you want to change (
NSRange), with the smaller/subscriptUIFont, and theNSBaselineOffsetAttributeNamevalue is the amount you want to offset it vertically. - Assign it to your
UILabel
- 得到
UIFont你想要的默认和上标样式，上标必须更小。 NSMutableAttributedString使用完整的字符串和默认字体创建一个。- 为要更改的字符 (
NSRange)添加一个属性，使用更小/下标UIFont，其NSBaselineOffsetAttributeName值是您要垂直偏移的量。 - 分配给你的
UILabel
Hopefully this helps other Swift devs as I needed this as well.
希望这可以帮助其他 Swift 开发人员，因为我也需要它。
回答by Chris
As a different approach, I wrote a function that takes in a string where the exponents are prepended with ^such as 2^2?3?5^2and returns 22?3?52
作为一种不同的方法，我编写了一个函数，该函数接受一个字符串，其中指数前面带有^例如2^2?3?5^2并返回22?3?52
func exponentize(str: String) -> String {
let supers = [
"1": "\u{00B9}",
"2": "\u{00B2}",
"3": "\u{00B3}",
"4": "\u{2074}",
"5": "\u{2075}",
"6": "\u{2076}",
"7": "\u{2077}",
"8": "\u{2078}",
"9": "\u{2079}"]
var newStr = ""
var isExp = false
for (_, char) in str.characters.enumerate() {
if char == "^" {
isExp = true
} else {
if isExp {
let key = String(char)
if supers.keys.contains(key) {
newStr.append(Character(supers[key]!))
} else {
isExp = false
newStr.append(char)
}
} else {
newStr.append(char)
}
}
}
return newStr
}
It's a bit of a brute force method, but it works if you don't want to deal with attributed strings or you want your string to be independent of a font.
这是一种蛮力方法，但如果您不想处理属性字符串或希望您的字符串独立于字体，它就可以工作。
回答by Atka
I wrote the following extension or you can use it as a function, it is working well for me . you can modify it by skipping the parts that are not essential to you
我编写了以下扩展，或者您可以将其用作函数，它对我来说效果很好。你可以通过跳过对你来说不重要的部分来修改它
extension NSMutableAttributedString
{
enum scripting : Int
{
case aSub = -1
case aSuper = 1
}
func characterSubscriptAndSuperscript(string:String,
characters:[Character],
type:scripting,
fontSize:CGFloat,
scriptFontSize:CGFloat,
offSet:Int,
length:[Int],
alignment:NSTextAlignment)-> NSMutableAttributedString
{
let paraghraphStyle = NSMutableParagraphStyle()
// Set The Paragraph aligmnet , you can ignore this part and delet off the function
paraghraphStyle.alignment = alignment
var scriptedCharaterLocation = Int()
//Define the fonts you want to use and sizes
let stringFont = UIFont.boldSystemFont(ofSize: fontSize)
let scriptFont = UIFont.boldSystemFont(ofSize: scriptFontSize)
// Define Attributes of the text body , this part can be removed of the function
let attString = NSMutableAttributedString(string:string, attributes: [NSFontAttributeName:stringFont,NSForegroundColorAttributeName:UIColor.black,NSParagraphStyleAttributeName: paraghraphStyle])
// the enum is used here declaring the required offset
let baseLineOffset = offSet * type.rawValue
// enumerated the main text characters using a for loop
for (i,c) in string.characters.enumerated()
{
// enumerated the array of first characters to subscript
for (theLength,aCharacter) in characters.enumerated()
{
if c == aCharacter
{
// Get to location of the first character
scriptedCharaterLocation = i
//Now set attributes starting from the character above
attString.setAttributes([NSFontAttributeName:scriptFont,
// baseline off set from . the enum i.e. +/- 1
NSBaselineOffsetAttributeName:baseLineOffset,
NSForegroundColorAttributeName:UIColor.black],
// the range from above location
range:NSRange(location:scriptedCharaterLocation,
// you define the length in the length array
// if subscripting at different location
// you need to define the length for each one
length:length[theLength]))
}
}
}
return attString}
}
examples:
例子：
let attStr1 = NSMutableAttributedString().characterSubscriptAndSuperscript(
string: "23 x 456",
characters:["3","5"],
type: .aSuper,
fontSize: 20,
scriptFontSize: 15,
offSet: 10,
length: [1,2],
alignment: .left)
let attStr2 = NSMutableAttributedString().characterSubscriptAndSuperscript(
string: "H2SO4",
characters: ["2","4"],
type: .aSub,
fontSize: 20,
scriptFontSize: 15,
offSet: 8,
length: [1,1],
alignment: .left)
回答by Glenn Howes
If you can get along with text that doesn't look perfect, and only need a subset of characters you can make use of the unicode superscript and subscript numbers: ? 1 2 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? This has the advantage of being a lot less cumbersome.
如果您可以处理看起来不完美的文本，并且只需要一个字符子集，您可以使用 unicode 上标和下标数字： ? 1 2 3 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 这具有不那么麻烦的优点。
回答by Jeehut
For a simple to use Swift solution, you might want to checkout HandyUIKit. After importing it into your project (e.g. via Carthage – see instructions in README) you can do something like this:
对于简单易用的 Swift 解决方案，您可能需要查看HandyUIKit。将其导入您的项目后（例如通过 Carthage – 请参阅自述文件中的说明），您可以执行以下操作：
import HandyUIKit
"6.022*10^{23}".superscripted(font: UIFont.systemFont(ofSize: 20, weight: .medium))
This line will return an NSAttributedStringwhich will look exactly like what you're looking for. Just assign itto a UILabels attributedTextproperty and that's it!
这一行将返回一个NSAttributedString看起来与您正在寻找的完全一样的。只需将其分配给UILabelsattributedText属性即可！
If you're looking for subscriptinga text, simply use subscripted(font:)instead. It will recognize structures like CO_{2}. There's also superAndSubscripted(font:)if you want to combine both.
如果您正在寻找下标文本，只需使用即可subscripted(font:)。它会识别像CO_{2}. 还有superAndSubscripted(font:)，如果你想结合两者。
See the docsfor more information and additional examples.
有关更多信息和其他示例，请参阅文档。
回答by SafeFastExpressive
Here is a simple version that has correct error handling and will compile in playground.
这是一个简单的版本，它具有正确的错误处理，并将在操场上编译。
import UIKit
func setMyLabelText(myLabel: UILabel) {
if let largeFont = UIFont(name: "Helvetica", size: 20), let superScriptFont = UIFont(name: "Helvetica", size:10) {
let numberString = NSMutableAttributedString(string: "6.022*10", attributes: [.font: largeFont])
numberString.append(NSAttributedString(string: "23", attributes: [.font: superScriptFont, .baselineOffset: 10]))
myLabel.attributedText = numberString
}
}
let myLabel = UILabel()
setMyLabelText(myLabel: myLabel)
回答by Abhishek Thapliyal
Swift 4+ Version of @Atka's Answer
@Atka 答案的 Swift 4+ 版本
import UIKit
extension NSMutableAttributedString {
enum Scripting : Int {
case aSub = -1
case aSuper = 1
}
func scripts(string: String,
characters: [Character],
type: Scripting,
stringFont: UIFont,
fontSize: CGFloat,
scriptFont: UIFont,
scriptFontSize: CGFloat,
offSet: Int,
length: [Int],
alignment: NSTextAlignment) -> NSMutableAttributedString {
let paraghraphStyle = NSMutableParagraphStyle()
paraghraphStyle.alignment = alignment
var scriptedCharaterLocation = Int()
let attributes = [
NSAttributedStringKey.font: stringFont,
NSAttributedStringKey.foregroundColor: UIColor.black,
NSAttributedStringKey.paragraphStyle: paraghraphStyle
]
let attString = NSMutableAttributedString(string:string, attributes: attributes)
let baseLineOffset = offSet * type.rawValue
let scriptTextAttributes: [NSAttributedStringKey : Any] = [
NSAttributedStringKey.font: scriptFont,
NSAttributedStringKey.baselineOffset: baseLineOffset,
NSAttributedStringKey.foregroundColor: UIColor.blue
]
for (i,c) in string.enumerated() {
for (theLength, aCharacter) in characters.enumerated() {
if c == aCharacter {
scriptedCharaterLocation = i
attString.setAttributes(scriptTextAttributes, range: NSRange(location:scriptedCharaterLocation,
length: length[theLength]))
}
}
}
return attString
}
}
回答by gorillaz
A nice simple function that outputs a number as the superscript text.
一个很好的简单函数，输出一个数字作为上标文本。
func exponent(i: Int) -> String {
let powers : [String] = [
"\u{2070}",
"\u{00B9}",
"\u{00B2}",
"\u{00B3}",
"\u{2074}",
"\u{2075}",
"\u{2076}",
"\u{2077}",
"\u{2078}",
"\u{2079}"
]
let digits = Array(String(i))
var string = ""
for d in digits {
string.append("\(powers[Int(String(d))!])")
}
return string
}
回答by henrik-dmg
I have created a String extension which takes a string and converts all of its superscript into unicode characters. This way you could for example share the resulting string without any hassle.
我创建了一个字符串扩展，它接受一个字符串并将其所有上标转换为 unicode 字符。例如，您可以通过这种方式轻松共享结果字符串。
extension Character {
var unicode: String {
// See table here: https://en.wikipedia.org/wiki/Unicode_subscripts_and_superscripts
let unicodeChars = [Character("0"):"\u{2070}",
Character("1"):"\u{00B9}",
Character("2"):"\u{00B2}",
Character("3"):"\u{00B3}",
Character("4"):"\u{2074}",
Character("5"):"\u{2075}",
Character("6"):"\u{2076}",
Character("7"):"\u{2077}",
Character("8"):"\u{2078}",
Character("9"):"\u{2079}",
Character("i"):"\u{2071}",
Character("+"):"\u{207A}",
Character("-"):"\u{207B}",
Character("="):"\u{207C}",
Character("("):"\u{207D}",
Character(")"):"\u{207E}",
Character("n"):"\u{207F}"]
if let unicode = unicodeChars[self] {
return unicode
}
return String(self)
}
}
extension String {
var unicodeSuperscript: String {
let char = Character(self)
return char.unicode
}
func superscripted() -> String {
let regex = try! NSRegularExpression(pattern: "\^\{([^\}]*)\}")
var unprocessedString = self
var resultString = String()
while let match = regex.firstMatch(in: unprocessedString, options: .reportCompletion, range: NSRange(location: 0, length: unprocessedString.count)) {
// add substring before match
let substringRange = unprocessedString.index(unprocessedString.startIndex, offsetBy: match.range.location)
let subString = unprocessedString.prefix(upTo: substringRange)
resultString.append(String(subString))
// add match with subscripted style
let capturedSubstring = NSAttributedString(string: unprocessedString).attributedSubstring(from: match.range(at: 1)).mutableCopy() as! NSMutableAttributedString
capturedSubstring.string.forEach { (char) in
let superScript = char.unicode
let string = NSAttributedString(string: superScript)
resultString.append(string.string)
}
// strip off the processed part
unprocessedString.deleteCharactersInRange(range: NSRange(location: 0, length: match.range.location + match.range.length))
}
// add substring after last match
resultString.append(unprocessedString)
return resultString
}
mutating func deleteCharactersInRange(range: NSRange) {
let mutableSelf = NSMutableString(string: self)
mutableSelf.deleteCharacters(in: range)
self = mutableSelf as String
}
}
For example "x^{4+n}+12^{3}".superscripted()produces "x???+123"
例如"x^{4+n}+12^{3}".superscripted()产生"x???+123"
This was inspired by HandyUIKitand the gist to my code is on Github
这受到HandyUIKit 的启发，我的代码的要点在Github 上
回答by Sajjon
Here's a Swift 5.1 solution (should work with older versions of Swift too) using recursion, that only focuses outputting a superscript from an Int(i.e. no formatting for display).
这是一个使用递归的 Swift 5.1 解决方案（也应该与旧版本的 Swift 一起使用），它只关注从 an 输出上标Int（即没有显示格式）。
extension Int {
func superscriptString() -> String {
let minusPrefixOrEmpty: String = self < 0 ? Superscript.minus : ""
let (quotient, remainder) = abs(self).quotientAndRemainder(dividingBy: 10)
let quotientString = quotient > 0 ? quotient.superscriptString() : ""
return minusPrefixOrEmpty + quotientString + Superscript.value(remainder)
}
}
enum Superscript {
static let minus = "?"
private static let values: [String] = [
"?",
"1",
"2",
"3",
"?",
"?",
"?",
"?",
"?",
"?"
]
static func value(_ int: Int) -> String {
assert(int >= 0 && int <= 9)
return values[int]
}
}
Here are some tests to prove correctness:
以下是一些证明正确性的测试：
func testPositiveIntegersSuperscript() {
XCTAssertEqual(0.superscriptString(), "?")
XCTAssertEqual(1.superscriptString(), "1")
XCTAssertEqual(2.superscriptString(), "2")
XCTAssertEqual(3.superscriptString(), "3")
XCTAssertEqual(4.superscriptString(), "?")
XCTAssertEqual(5.superscriptString(), "?")
XCTAssertEqual(6.superscriptString(), "?")
XCTAssertEqual(7.superscriptString(), "?")
XCTAssertEqual(8.superscriptString(), "?")
XCTAssertEqual(9.superscriptString(), "?")
XCTAssertEqual(10.superscriptString(), "1?")
XCTAssertEqual(11.superscriptString(), "11")
XCTAssertEqual(12.superscriptString(), "12")
XCTAssertEqual(19.superscriptString(), "1?")
XCTAssertEqual(20.superscriptString(), "2?")
XCTAssertEqual(21.superscriptString(), "21")
XCTAssertEqual(99.superscriptString(), "??")
XCTAssertEqual(100.superscriptString(), "1??")
XCTAssertEqual(101.superscriptString(), "1?1")
XCTAssertEqual(102.superscriptString(), "1?2")
XCTAssertEqual(237.superscriptString(), "23?")
XCTAssertEqual(999.superscriptString(), "???")
XCTAssertEqual(1000.superscriptString(), "1???")
XCTAssertEqual(1001.superscriptString(), "1??1")
XCTAssertEqual(1234.superscriptString(), "123?")
XCTAssertEqual(1337.superscriptString(), "133?")
}
func testNegativeIntegersSuperscript() {
XCTAssertEqual(Int(-1).superscriptString(), "?1")
XCTAssertEqual(Int(-2).superscriptString(), "?2")
XCTAssertEqual(Int(-3).superscriptString(), "?3")
XCTAssertEqual(Int(-4).superscriptString(), "??")
XCTAssertEqual(Int(-5).superscriptString(), "??")
XCTAssertEqual(Int(-6).superscriptString(), "??")
XCTAssertEqual(Int(-7).superscriptString(), "??")
XCTAssertEqual(Int(-8).superscriptString(), "??")
XCTAssertEqual(Int(-9).superscriptString(), "??")
XCTAssertEqual(Int(-10).superscriptString(), "?1?")
XCTAssertEqual(Int(-11).superscriptString(), "?11")
XCTAssertEqual(Int(-12).superscriptString(), "?12")
XCTAssertEqual(Int(-19).superscriptString(), "?1?")
XCTAssertEqual(Int(-20).superscriptString(), "?2?")
XCTAssertEqual(Int(-21).superscriptString(), "?21")
XCTAssertEqual(Int(-99).superscriptString(), "???")
XCTAssertEqual(Int(-100).superscriptString(), "?1??")
XCTAssertEqual(Int(-101).superscriptString(), "?1?1")
XCTAssertEqual(Int(-102).superscriptString(), "?1?2")
XCTAssertEqual(Int(-237).superscriptString(), "?23?")
XCTAssertEqual(Int(-999).superscriptString(), "????")
XCTAssertEqual(Int(-1000).superscriptString(), "?1???")
XCTAssertEqual(Int(-1001).superscriptString(), "?1??1")
XCTAssertEqual(Int(-1234).superscriptString(), "?123?")
XCTAssertEqual(Int(-1337).superscriptString(), "?133?")
}
My solution is more than twice as fast as gorillaz' solution(which is string and array based), thanks to mine being math and recursion based. Here is proof:
我的解决方案比gorillaz 的解决方案（基于字符串和数组）快两倍多，这要归功于我的基于数学和递归的解决方案。这是证明：
private typealias SuperscriptVector = (value: Int, expectedSuperstring: String)
private let vector1to9: SuperscriptVector = (123456789, "123??????")
func performanceTest(times n: Int, function: (Int) -> () -> String) {
func manyTimes(_ times: Int) {
func doTest(vector: SuperscriptVector) {
let result: String = function(vector.value)()
XCTAssertEqual(result, vector.expectedSuperstring)
}
for _ in 0..<times {
doTest(vector: vector1to9)
}
}
manyTimes(n)
}
// 3.244 sec
func testPerformanceMine() {
measure {
performanceTest(times: 1_000_000, function: Int.superscriptString)
}
}
// 7.6 sec
func testPerformanceStackOverflow() {
measure {
performanceTest(times: 1_000_000, function: Int.superscriptStringArrayBased)
}
}
