Ios 如何在swift中每N个字符向字符串添加分隔符?
我有一个包含二进制数字的字符串。如何将它分成几对数字 假设字符串为:Ios 如何在swift中每N个字符向字符串添加分隔符?,ios,string,swift,character,Ios,String,Swift,Character,我有一个包含二进制数字的字符串。如何将它分成几对数字 假设字符串为: let x = "11231245" 我想在每2个字符后添加一个分隔符,如“:”(即冒号) 我希望输出为: "11:23:12:45" 如何在Swift中执行此操作?Swift 5.2•Xcode 11.4或更高版本 extension Collection { func unfoldSubSequences(limitedTo maxLength: Int) -> UnfoldSequence<Sub
let x = "11231245"
我想在每2个字符后添加一个分隔符,如“:”(即冒号)
我希望输出为:
"11:23:12:45"
如何在Swift中执行此操作?Swift 5.2•Xcode 11.4或更高版本
extension Collection {
func unfoldSubSequences(limitedTo maxLength: Int) -> UnfoldSequence<SubSequence,Index> {
sequence(state: startIndex) { start in
guard start < self.endIndex else { return nil }
let end = self.index(start, offsetBy: maxLength, limitedBy: self.endIndex) ?? self.endIndex
defer { start = end }
return self[start..<end]
}
}
func every(n: Int) -> UnfoldSequence<Element,Index> {
sequence(state: startIndex) { index in
guard index < endIndex else { return nil }
defer { index = self.index(index, offsetBy: n, limitedBy: endIndex) ?? endIndex }
return self[index]
}
}
var pairs: [SubSequence] { .init(unfoldSubSequences(limitedTo: 2)) }
}
我对该代码的尝试是:
func insert(seperator: String, afterEveryXChars: Int, intoString: String) -> String {
var output = ""
intoString.characters.enumerate().forEach { index, c in
if index % afterEveryXChars == 0 && index > 0 {
output += seperator
}
output.append(c)
}
return output
}
insert(":", afterEveryXChars: 2, intoString: "11231245")
哪个输出
11:23:12:45
我的密码是swift 4
let x = "11231245"
var newText = String()
for (index, character) in x.enumerated() {
if index != 0 && index % 2 == 0 {
newText.append(":")
}
newText.append(String(character))
}
print(newText)
输出11:23:12:45简短而简单,如果需要,可以添加一个或两个
let
extension String {
func separate(every: Int, with separator: String) -> String {
return String(stride(from: 0, to: Array(self).count, by: every).map {
Array(Array(self)[$0..<min($0 + every, Array(self).count)])
}.joined(separator: separator))
}
}
扩展字符串{
func separate(every:Int,带分隔符:String)->String{
返回字符串(步幅(从:0到:数组(self).count,by:every).map{
数组(数组(self)[$0..我会选择这个紧凑的解决方案(在Swift 4中):
您可以进行扩展并参数化步幅和分隔符,以便可以对所需的每个值使用它(在我的示例中,我使用它转储32位空间操作的十六进制数据):
在您的情况下,这将给出以下结果:
let x = "11231245"
print (x.separate(every:2, with: ":")
$ 11:23:12:45
扩展字符串{
func separate(every:Int)->[字符串]{
返回步幅(从:0,到:计数,按:每个)。映射{
设ix0=索引(startIndex,offsetBy:$0);
设ix1=索引(在:ix0之后);
如果ix1 parts.append(String(self[ix0..一个简单的字符串扩展,它不要求原始字符串是步长(增量)的倍数):
Swift 4.2.1-Xcode 10.1
extension String {
func insertSeparator(_ separatorString: String, atEvery n: Int) -> String {
guard 0 < n else { return self }
return self.enumerated().map({String($0.element) + (($0.offset != self.count - 1 && $0.offset % n == n - 1) ? "\(separatorString)" : "")}).joined()
}
mutating func insertedSeparator(_ separatorString: String, atEvery n: Int) {
self = insertSeparator(separatorString, atEvery: n)
}
}
插入分隔符的简单单行代码(Swift 4.2):-
Swift 5.3
/// Adds a separator at every N characters
/// - Parameters:
/// - separator: the String value to be inserted, to separate the groups. Default is " " - one space.
/// - stride: the number of characters in the group, before a separator is inserted. Default is 4.
/// - Returns: Returns a String which includes a `separator` String at every `stride` number of characters.
func separated(by separator: String = " ", stride: Int = 4) -> String {
return enumerated().map { $0.isMultiple(of: stride) && ($0 != 0) ? "\(separator)\($1)" : String($1) }.joined()
}
我来晚了一点,但我喜欢这样使用正则表达式:
extension String {
func separating(every: Int, separator: String) -> String {
let regex = #"(.{\#(every)})(?=.)"#
return self.replacingOccurrences(of: regex, with: "$1\(separator)", options: [.regularExpression])
}
}
"111222333".separating(every: 3, separator: " ")
输出:
"111 222 333"
为了记录在案和您将来的利益,:
是一个“冒号”。“逗号”的意思是,
。到目前为止,您尝试了什么?请注意,您提供的唯一代码行实际上甚至都不是有效代码。您介意我编辑您的代码吗?不需要使用计数器,您可以使用enumerate()。您还可以添加一个附加条件(index>0)添加分隔符,以便能够将dropFirst方法删除为String.characters.enumerate().forEach{index,c in if index%afteryxChars==0&&index>0{…@LeoDabus一点也不,继续吧-你还在玩吗?:Pnot在玩。我对我的很满意:)为了让您知道如何使用enumerate(),代码会留下一个尾随:
,请尝试将:
添加到$0.element
前面,当且仅当您不在第一个字符处时。只需添加.flatte().dropLast()请求“在每两个字符之后”。如果尾随“:不需要,是的,您的建议会这样做,即:$0.index>0&&$0.index%2==0?[“:”,$0.element]:[$0.element],
@LeoDabus,不,如果原始字符串长度为奇数,dropLast将使用有效负载字符。@AntonBronnikov字符串(x.characters.enumerate().map(){$0.index%2==1?[$0.element]:[“:”,$0.element],$0.element]}.flant().dropFirst()@LeoDabus您编写了这些扩展吗?如果是,您能帮我做些什么吗?我正在尝试插入一个“|”每70个字符,但仅当第70个字符是空格时。如果不是,则向后跨步,直到它找到空格。我基本上将其用作一个新的行生成器。我似乎不知道如何执行此操作,我尝试了许多不同的方法。是的,我写了所有这些方法。请随意发布您解决问题的方法以及出现的问题ng在一个新问题中,我会看一看。@LeoDabus很好,在这里看一看。@TheValyreanGroup所以你想在9号之后拆分字符串的每个空格?哦,是的……太简单了。我太笨了。非常感谢你的帮助。祝你过得愉快!我的xcode在声明函数时要求返回,类似这样:func separate(每一步:Int=4,带分隔符:Character=“”)->String{这不可编译。@TruMan1:您使用的是什么Xcode和Swift版本?抱歉,我刚刚意识到我试图使用字符串作为分隔符。我喜欢您是一行,实际上更容易理解(我不知道性能比较如何)。如何调整扩展名以接受字符串作为分隔符?当我切换到字符串类型时,我遇到的错误是:无法使用类型为“(FlattCollection)”@StéphanedeLuca的参数列表调用类型为“init(:)”的初始值设定项。您只需使用flatMap而不是map并加入.init(enumerated().flatMap即可{$0>0&&$0%stride==0?[separator,$1]:[$1]}
newText.append(character)
这是一个非常好的解决方案。我认为这是最快捷、最容易理解的解决方案。谢谢。我添加了一个小补丁。这是一个额外的检查-因为我为第一个字符添加了额外的空格(0是任意数字的倍数)。
let x = "11231245"
print (x.separate(every:2, with: ":")
$ 11:23:12:45
extension String{
func separate(every: Int) -> [String] {
return stride(from: 0, to: count, by: every).map {
let ix0 = index(startIndex, offsetBy: $0);
let ix1 = index(after:ix0);
if ix1 < endIndex {
return String(self[ix0...ix1]);
}else{
return String(self[ix0..<endIndex]);
}
}
}
func separate(every: Int) -> [String] {
var parts:[String] = [];
var ix1 = startIndex;
while ix1 < endIndex {
let ix0 = ix1;
var n = 0;
while ix1 < endIndex && n < every {
ix1 = index(after: ix1);
n += 1;
}
parts.append(String(self[ix0..<ix1]));
}
return parts;
}
"asdf234sdf".separate(every: 2).joined(separator: ":");
extension String {
func inserted(_ newElement: Character,atEach increment:Int)->String {
var newStr = self
for indx in stride(from: increment, to: newStr.count, by: increment).reversed() {
let index = String.Index(encodedOffset: indx)
newStr.insert(newElement, at: index)
}
return newStr
}
}
extension String {
func insertSeparator(_ separatorString: String, atEvery n: Int) -> String {
guard 0 < n else { return self }
return self.enumerated().map({String($0.element) + (($0.offset != self.count - 1 && $0.offset % n == n - 1) ? "\(separatorString)" : "")}).joined()
}
mutating func insertedSeparator(_ separatorString: String, atEvery n: Int) {
self = insertSeparator(separatorString, atEvery: n)
}
}
let testString = "11231245"
let test1 = testString.insertSeparator(":", atEvery: 2)
print(test1) // 11:23:12:45
var test2 = testString
test2.insertedSeparator(",", atEvery: 3)
print(test2) // 112,312,45
let testString = "123456789"
let ansTest = testString.enumerated().compactMap({ ($0 > 0) && ($0 % 2 == 0) ? ":\($1)" : "\($1)" }).joined() ?? ""
print(ansTest) // 12:34:56:78:9
/// Adds a separator at every N characters
/// - Parameters:
/// - separator: the String value to be inserted, to separate the groups. Default is " " - one space.
/// - stride: the number of characters in the group, before a separator is inserted. Default is 4.
/// - Returns: Returns a String which includes a `separator` String at every `stride` number of characters.
func separated(by separator: String = " ", stride: Int = 4) -> String {
return enumerated().map { $0.isMultiple(of: stride) && ($0 != 0) ? "\(separator)\($1)" : String($1) }.joined()
}
extension String {
func separating(every: Int, separator: String) -> String {
let regex = #"(.{\#(every)})(?=.)"#
return self.replacingOccurrences(of: regex, with: "$1\(separator)", options: [.regularExpression])
}
}
"111222333".separating(every: 3, separator: " ")
"111 222 333"