I need to validate password entered by user and check if the password contains at least one uppercase and one lowercase char in Dart.
I wrote this String extension:
extension StringValidators on String {
bool containsUppercase() {
// What code should be here?
}
bool containsLowercase() {
// What code should be here?
}
}
And use it like this:
final text = passwordTextController.text;
final isValid = text.containsUppercase() && text.containsLowercase();
Is there any regexp for this purpose? Or it should be plain algorithm? Please help me to find out the elegant way. Thanks!
Minimum 1 Upper case,
Minimum 1 lowercase,
Minimum 1 Numeric Number,
Minimum 1 Special Character,
Common Allow Character ( ! # # $ & * ~ )
bool validateStructure(String value){
String pattern = r'^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[!##\$&*~]).{8,}$';
RegExp regExp = new RegExp(pattern);
return regExp.hasMatch(value);
}
extension StringValidators on String {
bool get containsUppercase => contains(RegExp(r'[A-Z]'));
bool get containsLowercase => contains(RegExp(r'[a-z]'));
}
For only minimum 1 upper and minimum 1 Lower only, you could use this RegEx:
RegExp regEx = new RegExp(r"(?=.*[a-z])(?=.*[A-Z])\w+");
String a = "aBc";
String b = "abc";
String c = "ABC";
print("a => " + regEx.hasMatch(a).toString());
print("b => " + regEx.hasMatch(b).toString());
print("c => " + regEx.hasMatch(c).toString());
Expected Result:
I/flutter (10220): a => true
I/flutter (10220): b => false
I/flutter (10220): c => false
Reusable
extension StringValidators on String {
meetsPasswordRequirements() {
RegExp regEx = new RegExp(r"(?=.*[a-z])(?=.*[A-Z])\w+");
return regEx.hasMatch(this);
}
}
Use
final isValid = text.meetsPasswordRequirements();
void main() {
solve("coDE");
}
String solve(String s) {
// your code here
List _a = s.split("");
String _b = "";
List _x = [];
List _y = [];
for(var i in _a){
if(i.toString() == i.toString().toUpperCase()){
_x.add(i);
}else{
_y.add(i);
}
}
if(_x.length == _y.length){
_b = _a.join().toLowerCase();
}else if(_x.length > _y.length){
_b = _a.join().toUpperCase();
}else if(_x.length < _y.length){
_b = _a.join().toLowerCase();
}
return "$_b";
}
OR
String solve2(String str) {
return RegExp(r'[A-Z]').allMatches(str).length >
RegExp(r'[a-z]').allMatches(str).length
? str.toUpperCase()
: str.toLowerCase();
}
Hello I'm new to Swift and am using SwiftUI for my project where I download some weather data and I display it in the ContentView().
I would like to highlight some part of the Text if it contains some specific word, but I don't have any idea how to start.
In ContentView(), I have tried to set a function receiving the string downloaded from web and return a string. I believe this is wrong, because SwiftUI does not apply the modifiers at the all for the Text.
For example, in my ContentView() I would like the word thunderstorm to have the .bold() modifier:
struct ContentView: View {
let testo : String = "There is a thunderstorm in the area"
var body: some View {
Text(highlight(str: testo))
}
func highlight(str: String) -> String {
let textToSearch = "thunderstorm"
var result = ""
if str.contains(textToSearch) {
let index = str.startIndex
result = String( str[index])
}
return result
}
}
If that requires just simple word styling then here is possible solution.
Tested with Xcode 11.4 / iOS 13.4
struct ContentView: View {
let testo : String = "There is a thunderstorm in the area. Added some testing long text to demo that wrapping works correctly!"
var body: some View {
hilightedText(str: testo, searched: "thunderstorm")
.multilineTextAlignment(.leading)
}
func hilightedText(str: String, searched: String) -> Text {
guard !str.isEmpty && !searched.isEmpty else { return Text(str) }
var result: Text!
let parts = str.components(separatedBy: searched)
for i in parts.indices {
result = (result == nil ? Text(parts[i]) : result + Text(parts[i]))
if i != parts.count - 1 {
result = result + Text(searched).bold()
}
}
return result ?? Text(str)
}
}
Note: below is previously used function, but as commented by #Lkabo it has limitations on very long strings
func hilightedText(str: String) -> Text {
let textToSearch = "thunderstorm"
var result: Text!
for word in str.split(separator: " ") {
var text = Text(word)
if word == textToSearch {
text = text.bold()
}
result = (result == nil ? text : result + Text(" ") + text)
}
return result ?? Text(str)
}
iOS 13, Swift 5. There is a generic solution described in this medium article. Using it you can highlight any text anywhere with the only catch being it cannot be more then 64 characters in length, since it using bitwise masks.
https://medium.com/#marklucking/an-interesting-challenge-with-swiftui-9ebb26e77376
This is the basic code in the article.
ForEach((0 ..< letter.count), id: \.self) { column in
Text(letter[column])
.foregroundColor(colorCode(gate: Int(self.gate), no: column) ? Color.black: Color.red)
.font(Fonts.futuraCondensedMedium(size: fontSize))
}
And this one to mask the text...
func colorCode(gate:Int, no:Int) -> Bool {
let bgr = String(gate, radix:2).pad(with: "0", toLength: 16)
let bcr = String(no, radix:2).pad(with: "0", toLength: 16)
let binaryColumn = 1 << no - 1
let value = UInt64(gate) & UInt64(binaryColumn)
let vr = String(value, radix:2).pad(with: "0", toLength: 16)
print("bg ",bgr," bc ",bcr,vr)
return value > 0 ? true:false
}
You can concatenate with multiple Text Views.
import SwiftUI
import PlaygroundSupport
struct ContentView: View {
var body: some View{
let testo : String = "There is a thunderstorm in the area"
let stringArray = testo.components(separatedBy: " ")
let stringToTextView = stringArray.reduce(Text(""), {
if $1 == "thunderstorm" {
return $0 + Text($1).bold() + Text(" ")
} else {
return $0 + Text($1) + Text(" ")
}
})
return stringToTextView
}
}
PlaygroundPage.current.setLiveView(ContentView())
If you are targeting iOS15 / macOS12 and above, you can use AttributedString. For example:
private struct HighlightedText: View {
let text: String
let highlighted: String
var body: some View {
Text(attributedString)
}
private var attributedString: AttributedString {
var attributedString = AttributedString(text)
if let range = attributedString.range(of: highlighted)) {
attributedString[range].backgroundColor = .yellow
}
return attributedString
}
}
If you want your match to be case insensitive, you could replace the line
if let range = attributedString.range(of: highlighted)
with
if let range = AttributedString(text.lowercased()).range(of: highlighted.lowercased())
The answer of #Asperi works well. Here is a modified variant with a search by array of single words:
func highlightedText(str: String, searched: [String]) -> Text {
guard !str.isEmpty && !searched.isEmpty else { return Text(str) }
var result: Text!
let parts = str.components(separatedBy: " ")
for part_index in parts.indices {
result = (result == nil ? Text("") : result + Text(" "))
if searched.contains(parts[part_index].trimmingCharacters(in: .punctuationCharacters)) {
result = result + Text(parts[part_index])
.bold()
.foregroundColor(.red)
}
else {
result = result + Text(parts[part_index])
}
}
return result ?? Text(str)
}
Usage example:
let str: String = "There is a thunderstorm in the area. Added some testing long text to demo that wrapping works correctly!"
let searched: [String] = ["thunderstorm", "wrapping"]
highlightedText(str: str, searched: searched)
.padding()
.background(.yellow)
You can also make AttributedString with markdown this way
do {
return try AttributedString(markdown: foreignSentence.replacing(word.foreign, with: "**\(word.foreign)**"))
} catch {
return AttributedString(foreignSentence)
}
and just use Text
Text(foreignSentenceMarkdown)
In a program I'm writing, I need to check to see if a character is a space (" "). Currently have this as the conditional but it's not working. Any ideas? Thanks in advance.
for(var k = indexOfCharBeingExamined; k < lineBeingExaminedChars.count; k++){
let charBeingExamined = lineBeingExaminedChars[lineBeingExaminedChars.startIndex.advancedBy(k)];
//operations
if(String(charBeingExamined) == " "){
//more operations
}
}
The code below is how I solved this problem with a functional approach in Swift. I made an extension (two, actually), but you could easily take the guts of the function and use it elsewhere.
extension String {
var isWhitespace: Bool {
guard !isEmpty else { return true }
let whitespaceChars = NSCharacterSet.whitespacesAndNewlines
return self.unicodeScalars
.filter { !whitespaceChars.contains($0) }
.count == 0
}
}
extension Optional where Wrapped == String {
var isNullOrWhitespace: Bool {
return self?.isWhitespace ?? true
}
}
The following code works for me. Note that it's easier to just iterate over the characters in a string using 'for' (second example below):
var s = "X yz"
for var i = 0; i < s.characters.count; i++ {
let x = s[s.startIndex.advancedBy(i)]
print(x)
print(String(x) == " ")
}
for c in s.characters {
print(c)
print(String(c) == " ")
}
String:
let origin = "Some string with\u{00a0}whitespaces" // \u{00a0} is a no-break space
Oneliner:
let result = origin.characters.contains { " \u{00a0}".characters.contains($0) }
Another approach:
let spaces = NSCharacterSet.whitespaceCharacterSet()
let result = origin.utf16.contains { spaces.characterIsMember($0) }
Output:
print(result) // true
Not sure what you want to do with the spaces, because then it could be a bit simpler.
just in your code change " " -> "\u{00A0}"
for(var k = indexOfCharBeingExamined; k < lineBeingExaminedChars.count; k++){
let charBeingExamined = lineBeingExaminedChars[lineBeingExaminedChars.startIndex.advancedBy(k)];
if(String(charBeingExamined) == "\u{00A0}"){
//more operations
}
}
To test just for whitespace:
func hasWhitespace(_ input: String) -> Bool {
let inputCharacterSet = CharacterSet(charactersIn: input)
return !inputCharacterSet.intersection(CharacterSet.whitespaces).isEmpty
}
To test for both whitespace and an empty string:
func hasWhitespace(_ input: String) -> Bool {
let inputCharacterSet = CharacterSet(charactersIn: input)
return !inputCharacterSet.intersection(CharacterSet.whitespaces).isEmpty || inputCharacterSet.isEmpty
}
It's time to admit defeat...
In Objective-C, I could use something like:
NSString* str = #"abcdefghi";
[str rangeOfString:#"c"].location; // 2
In Swift, I see something similar:
var str = "abcdefghi"
str.rangeOfString("c").startIndex
...but that just gives me a String.Index, which I can use to subscript back into the original string, but not extract a location from.
FWIW, that String.Index has a private ivar called _position that has the correct value in it. I just don't see how it's exposed.
I know I could easily add this to String myself. I'm more curious about what I'm missing in this new API.
You are not the only one who couldn't find the solution.
String doesn't implement RandomAccessIndexType. Probably because they enable characters with different byte lengths. That's why we have to use string.characters.count (count or countElements in Swift 1.x) to get the number of characters. That also applies to positions. The _position is probably an index into the raw array of bytes and they don't want to expose that. The String.Index is meant to protect us from accessing bytes in the middle of characters.
That means that any index you get must be created from String.startIndex or String.endIndex (String.Index implements BidirectionalIndexType). Any other indices can be created using successor or predecessor methods.
Now to help us with indices, there is a set of methods (functions in Swift 1.x):
Swift 4.x
let text = "abc"
let index2 = text.index(text.startIndex, offsetBy: 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let characterIndex2 = text.index(text.startIndex, offsetBy: 2)
let lastChar2 = text[characterIndex2] //will do the same as above
let range: Range<String.Index> = text.range(of: "b")!
let index: Int = text.distance(from: text.startIndex, to: range.lowerBound)
Swift 3.0
let text = "abc"
let index2 = text.index(text.startIndex, offsetBy: 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let characterIndex2 = text.characters.index(text.characters.startIndex, offsetBy: 2)
let lastChar2 = text.characters[characterIndex2] //will do the same as above
let range: Range<String.Index> = text.range(of: "b")!
let index: Int = text.distance(from: text.startIndex, to: range.lowerBound)
Swift 2.x
let text = "abc"
let index2 = text.startIndex.advancedBy(2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let lastChar2 = text.characters[index2] //will do the same as above
let range: Range<String.Index> = text.rangeOfString("b")!
let index: Int = text.startIndex.distanceTo(range.startIndex) //will call successor/predecessor several times until the indices match
Swift 1.x
let text = "abc"
let index2 = advance(text.startIndex, 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let range = text.rangeOfString("b")
let index: Int = distance(text.startIndex, range.startIndex) //will call succ/pred several times
Working with String.Index is cumbersome but using a wrapper to index by integers (see https://stackoverflow.com/a/25152652/669586) is dangerous because it hides the inefficiency of real indexing.
Note that Swift indexing implementation has the problem that indices/ranges created for one string cannot be reliably used for a different string, for example:
Swift 2.x
let text: String = "abc"
let text2: String = "πΎππ"
let range = text.rangeOfString("b")!
//can randomly return a bad substring or throw an exception
let substring: String = text2[range]
//the correct solution
let intIndex: Int = text.startIndex.distanceTo(range.startIndex)
let startIndex2 = text2.startIndex.advancedBy(intIndex)
let range2 = startIndex2...startIndex2
let substring: String = text2[range2]
Swift 1.x
let text: String = "abc"
let text2: String = "πΎππ"
let range = text.rangeOfString("b")
//can randomly return nil or a bad substring
let substring: String = text2[range]
//the correct solution
let intIndex: Int = distance(text.startIndex, range.startIndex)
let startIndex2 = advance(text2.startIndex, intIndex)
let range2 = startIndex2...startIndex2
let substring: String = text2[range2]
Swift 3.0 makes this a bit more verbose:
let string = "Hello.World"
let needle: Character = "."
if let idx = string.characters.index(of: needle) {
let pos = string.characters.distance(from: string.startIndex, to: idx)
print("Found \(needle) at position \(pos)")
}
else {
print("Not found")
}
Extension:
extension String {
public func index(of char: Character) -> Int? {
if let idx = characters.index(of: char) {
return characters.distance(from: startIndex, to: idx)
}
return nil
}
}
In Swift 2.0 this has become easier:
let string = "Hello.World"
let needle: Character = "."
if let idx = string.characters.indexOf(needle) {
let pos = string.startIndex.distanceTo(idx)
print("Found \(needle) at position \(pos)")
}
else {
print("Not found")
}
Extension:
extension String {
public func indexOfCharacter(char: Character) -> Int? {
if let idx = self.characters.indexOf(char) {
return self.startIndex.distanceTo(idx)
}
return nil
}
}
Swift 1.x implementation:
For a pure Swift solution one can use:
let string = "Hello.World"
let needle: Character = "."
if let idx = find(string, needle) {
let pos = distance(string.startIndex, idx)
println("Found \(needle) at position \(pos)")
}
else {
println("Not found")
}
As an extension to String:
extension String {
public func indexOfCharacter(char: Character) -> Int? {
if let idx = find(self, char) {
return distance(self.startIndex, idx)
}
return nil
}
}
Swift 5.0
public extension String {
func indexInt(of char: Character) -> Int? {
return firstIndex(of: char)?.utf16Offset(in: self)
}
}
Swift 4.0
public extension String {
func indexInt(of char: Character) -> Int? {
return index(of: char)?.encodedOffset
}
}
extension String {
// MARK: - sub String
func substringToIndex(index:Int) -> String {
return self.substringToIndex(advance(self.startIndex, index))
}
func substringFromIndex(index:Int) -> String {
return self.substringFromIndex(advance(self.startIndex, index))
}
func substringWithRange(range:Range<Int>) -> String {
let start = advance(self.startIndex, range.startIndex)
let end = advance(self.startIndex, range.endIndex)
return self.substringWithRange(start..<end)
}
subscript(index:Int) -> Character{
return self[advance(self.startIndex, index)]
}
subscript(range:Range<Int>) -> String {
let start = advance(self.startIndex, range.startIndex)
let end = advance(self.startIndex, range.endIndex)
return self[start..<end]
}
// MARK: - replace
func replaceCharactersInRange(range:Range<Int>, withString: String!) -> String {
var result:NSMutableString = NSMutableString(string: self)
result.replaceCharactersInRange(NSRange(range), withString: withString)
return result
}
}
I have found this solution for swift2:
var str = "abcdefghi"
let indexForCharacterInString = str.characters.indexOf("c") //returns 2
I'm not sure how to extract the position from String.Index, but if you're willing to fall back on some Objective-C frameworks, you can bridge to objective-c and do it the same way you used to.
"abcdefghi".bridgeToObjectiveC().rangeOfString("c").location
It seems like some NSString methods haven't yet been (or maybe won't be) ported to String. Contains also comes to mind.
Here is a clean String extention that answers the question:
Swift 3:
extension String {
var length:Int {
return self.characters.count
}
func indexOf(target: String) -> Int? {
let range = (self as NSString).range(of: target)
guard range.toRange() != nil else {
return nil
}
return range.location
}
func lastIndexOf(target: String) -> Int? {
let range = (self as NSString).range(of: target, options: NSString.CompareOptions.backwards)
guard range.toRange() != nil else {
return nil
}
return self.length - range.location - 1
}
func contains(s: String) -> Bool {
return (self.range(of: s) != nil) ? true : false
}
}
Swift 2.2:
extension String {
var length:Int {
return self.characters.count
}
func indexOf(target: String) -> Int? {
let range = (self as NSString).rangeOfString(target)
guard range.toRange() != nil else {
return nil
}
return range.location
}
func lastIndexOf(target: String) -> Int? {
let range = (self as NSString).rangeOfString(target, options: NSStringCompareOptions.BackwardsSearch)
guard range.toRange() != nil else {
return nil
}
return self.length - range.location - 1
}
func contains(s: String) -> Bool {
return (self.rangeOfString(s) != nil) ? true : false
}
}
You can also find indexes of a character in a single string like this,
extension String {
func indexes(of character: String) -> [Int] {
precondition(character.count == 1, "Must be single character")
return self.enumerated().reduce([]) { partial, element in
if String(element.element) == character {
return partial + [element.offset]
}
return partial
}
}
}
Which gives the result in [String.Distance] ie. [Int], like
"apple".indexes(of: "p") // [1, 2]
"element".indexes(of: "e") // [0, 2, 4]
"swift".indexes(of: "j") // []
Swift 5
Find index of substring
let str = "abcdecd"
if let range: Range<String.Index> = str.range(of: "cd") {
let index: Int = str.distance(from: str.startIndex, to: range.lowerBound)
print("index: ", index) //index: 2
}
else {
print("substring not found")
}
Find index of Character
let str = "abcdecd"
if let firstIndex = str.firstIndex(of: "c") {
let index: Int = str.distance(from: str.startIndex, to: firstIndex)
print("index: ", index) //index: 2
}
else {
print("symbol not found")
}
If you want to use familiar NSString, you can declare it explicitly:
var someString: NSString = "abcdefghi"
var someRange: NSRange = someString.rangeOfString("c")
I'm not sure yet how to do this in Swift.
If you want to know the position of a character in a string as an int value use this:
let loc = newString.range(of: ".").location
This worked for me,
var loc = "abcdefghi".rangeOfString("c").location
NSLog("%d", loc);
this worked too,
var myRange: NSRange = "abcdefghi".rangeOfString("c")
var loc = myRange.location
NSLog("%d", loc);
I know this is old and an answer has been accepted, but you can find the index of the string in a couple lines of code using:
var str : String = "abcdefghi"
let characterToFind: Character = "c"
let characterIndex = find(str, characterToFind) //returns 2
Some other great information about Swift strings here Strings in Swift
Variable type String in Swift contains different functions compared to NSString in Objective-C . And as Sulthan mentioned,
Swift String doesn't implement RandomAccessIndex
What you can do is downcast your variable of type String to NSString (this is valid in Swift). This will give you access to the functions in NSString.
var str = "abcdefghi" as NSString
str.rangeOfString("c").locationx // returns 2
If you think about it, you actually don't really need the exact Int version of the location. The Range or even the String.Index is enough to get the substring out again if needed:
let myString = "hello"
let rangeOfE = myString.rangeOfString("e")
if let rangeOfE = rangeOfE {
myString.substringWithRange(rangeOfE) // e
myString[rangeOfE] // e
// if you do want to create your own range
// you can keep the index as a String.Index type
let index = rangeOfE.startIndex
myString.substringWithRange(Range<String.Index>(start: index, end: advance(index, 1))) // e
// if you really really need the
// Int version of the index:
let numericIndex = distance(index, advance(index, 1)) // 1 (type Int)
}
The Simplest Way is:
In Swift 3:
var textViewString:String = "HelloWorld2016"
guard let index = textViewString.characters.index(of: "W") else { return }
let mentionPosition = textViewString.distance(from: index, to: textViewString.endIndex)
print(mentionPosition)
String is a bridge type for NSString, so add
import Cocoa
to your swift file and use all the "old" methods.
In terms of thinking this might be called an INVERSION. You discover the world is round instead of flat. "You don't really need to know the INDEX of the character to do things with it." And as a C programmer I found that hard to take too!
Your line "let index = letters.characters.indexOf("c")!" is enough by itself.
For example to remove the c you could use...(playground paste in)
var letters = "abcdefg"
//let index = letters.rangeOfString("c")!.startIndex //is the same as
let index = letters.characters.indexOf("c")!
range = letters.characters.indexOf("c")!...letters.characters.indexOf("c")!
letters.removeRange(range)
letters
However, if you want an index you need to return an actual INDEX not an Int as an Int value would require additional steps for any practical use. These extensions return an index, a count of a specific character, and a range which this playground plug-in-able code will demonstrate.
extension String
{
public func firstIndexOfCharacter(aCharacter: Character) -> String.CharacterView.Index? {
for index in self.characters.indices {
if self[index] == aCharacter {
return index
}
}
return nil
}
public func returnCountOfThisCharacterInString(aCharacter: Character) -> Int? {
var count = 0
for letters in self.characters{
if aCharacter == letters{
count++
}
}
return count
}
public func rangeToCharacterFromStart(aCharacter: Character) -> Range<Index>? {
for index in self.characters.indices {
if self[index] == aCharacter {
let range = self.startIndex...index
return range
}
}
return nil
}
}
var MyLittleString = "MyVery:important String"
var theIndex = MyLittleString.firstIndexOfCharacter(":")
var countOfColons = MyLittleString.returnCountOfThisCharacterInString(":")
var theCharacterAtIndex:Character = MyLittleString[theIndex!]
var theRange = MyLittleString.rangeToCharacterFromStart(":")
MyLittleString.removeRange(theRange!)
Swift 4 Complete Solution:
OffsetIndexableCollection (String using Int Index)
https://github.com/frogcjn/OffsetIndexableCollection-String-Int-Indexable-
let a = "01234"
print(a[0]) // 0
print(a[0...4]) // 01234
print(a[...]) // 01234
print(a[..<2]) // 01
print(a[...2]) // 012
print(a[2...]) // 234
print(a[2...3]) // 23
print(a[2...2]) // 2
if let number = a.index(of: "1") {
print(number) // 1
print(a[number...]) // 1234
}
if let number = a.index(where: { $0 > "1" }) {
print(number) // 2
}
extension String {
//Fucntion to get the index of a particular string
func index(of target: String) -> Int? {
if let range = self.range(of: target) {
return characters.distance(from: startIndex, to: range.lowerBound)
} else {
return nil
}
}
//Fucntion to get the last index of occurence of a given string
func lastIndex(of target: String) -> Int? {
if let range = self.range(of: target, options: .backwards) {
return characters.distance(from: startIndex, to: range.lowerBound)
} else {
return nil
}
}
}
You can find the index number of a character in a string with this:
var str = "abcdefghi"
if let index = str.firstIndex(of: "c") {
let distance = str.distance(from: str.startIndex, to: index)
// distance is 2
}
If you are looking for easy way to get index of Character or String checkout this library http://www.dollarswift.org/#indexof-char-character-int
You can get the indexOf from a string using another string as well or regex pattern
To get index of a substring in a string with Swift 2:
let text = "abc"
if let range = text.rangeOfString("b") {
var index: Int = text.startIndex.distanceTo(range.startIndex)
...
}
In swift 2.0
var stringMe="Something In this.World"
var needle="."
if let idx = stringMe.characters.indexOf(needle) {
let pos=stringMe.substringFromIndex(idx)
print("Found \(needle) at position \(pos)")
}
else {
print("Not found")
}
let mystring:String = "indeep";
let findCharacter:Character = "d";
if (mystring.characters.contains(findCharacter))
{
let position = mystring.characters.indexOf(findCharacter);
NSLog("Position of c is \(mystring.startIndex.distanceTo(position!))")
}
else
{
NSLog("Position of c is not found");
}
I play with following
extension String {
func allCharactes() -> [Character] {
var result: [Character] = []
for c in self.characters {
result.append(c)
}
return
}
}
until I understand the provided one's now it's just Character array
and with
let c = Array(str.characters)
If you only need the index of a character the most simple, quick solution (as already pointed out by Pascal) is:
let index = string.characters.index(of: ".")
let intIndex = string.distance(from: string.startIndex, to: index)
On the subject of turning a String.Index into an Int, this extension works for me:
public extension Int {
/// Creates an `Int` from a given index in a given string
///
/// - Parameters:
/// - index: The index to convert to an `Int`
/// - string: The string from which `index` came
init(_ index: String.Index, in string: String) {
self.init(string.distance(from: string.startIndex, to: index))
}
}
Example usage relevant to this question:
var testString = "abcdefg"
Int(testString.range(of: "c")!.lowerBound, in: testString) // 2
testString = "π¨π¦πΊπΈπ©πͺπ©βπ©βπ§βπ¦\u{1112}\u{1161}\u{11AB}"
Int(testString.range(of: "π¨π¦πΊπΈπ©πͺ")!.lowerBound, in: testString) // 0
Int(testString.range(of: "π©βπ©βπ§βπ¦")!.lowerBound, in: testString) // 1
Int(testString.range(of: "αα
‘α«")!.lowerBound, in: testString) // 5
Important:
As you can tell, it groups extended grapheme clusters and joined characters differently than String.Index. Of course, this is why we have String.Index. You should keep in mind that this method considers clusters to be singular characters, which is closer to correct. If your goal is to split a string by Unicode codepoint, this is not the solution for you.
In Swift 2.0, the following function returns a substring before a given character.
func substring(before sub: String) -> String {
if let range = self.rangeOfString(sub),
let index: Int = self.startIndex.distanceTo(range.startIndex) {
return sub_range(0, index)
}
return ""
}
As my perspective, The better way with knowing the logic itself is below
let testStr: String = "I love my family if you Love us to tell us I'm with you"
var newStr = ""
let char:Character = "i"
for value in testStr {
if value == char {
newStr = newStr + String(value)
}
}
print(newStr.count)