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If String a = "abbc" and String b="abc", we have to print that character 'b' is missing in the second string.
I want to do it by using Java. I am able to do it when String 2 has a character not present in String 1 when s1=abc and s2=abk but not when characters are same in both strings like the one I have mentioned in the question.
public class Program
{
public static void main(String[] args) {
String str1 = "abbc";
String str2 = "abc";
char first[] = str1.toCharArray();
char second[] = str2.toCharArray();
HashMap <Character, Integer> map1 = new HashMap<Character,Integer>();
for(char a: first){
if(!map1.containsKey(a)){
map1.put(a,1);
}else{
map1.put(a,map1.get(a)+1);
}
}
System.out.println(map1);
HashMap <Character, Integer> map2 = new HashMap<Character,Integer>();
for(char b: second){
if(!map2.containsKey(b)){
map2.put(b,1);
}else{
map2.put(b,map2.get(b)+1);
}
}
System.out.println(map2);
}
}
I have two hashmaps here one for the longer string and one for the shorter string, map1 {a=1,b=2,c=1} and map2 {a=1,b=1,c=1}. What should I do after this?
Let assume that we have two strings a and b.
(optional) Compare lengths to find longer one.
Iterate over them char by char and compare letters at same index.
If both letters are the same, ignore it. If different, add letter from longer string to result and increment index of the longer string by 1.
What's left in longer string is your result.
Pseudocode:
const a = "aabbccc"
const b = "aabcc"
let res = ""
for (let i = 0, j = 0; i <= a.length; i++, j++) {
if (a[i] !== b[j]) {
res += a[i]
i++
}
}
console.log(res)
More modern and elegant way using high order functions:
const a = "aabbccc"
const b = "aabcc"
const res = [...a].reduce((r, e, i) => e === b[i - r.length] ? r : r + e, "")
console.log(res)
I need to extract numbers from string and put them into a new array in Swift.
var str = "I have to buy 3 apples, 7 bananas, 10eggs"
I tried to loop each characters and I have no idea to compare between Characters and Int.
Swift 3/4
let string = "0kaksd020dk2kfj2123"
if let number = Int(string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()) {
// Do something with this number
}
You can also make an extension like:
extension Int {
static func parse(from string: String) -> Int? {
return Int(string.components(separatedBy: CharacterSet.decimalDigits.inverted).joined())
}
}
And then later use it like:
if let number = Int.parse(from: "0kaksd020dk2kfj2123") {
// Do something with this number
}
First, we split the string so we can process the single items. Then we use NSCharacterSet to select the numbers only.
import Foundation
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let strArr = str.split(separator: " ")
for item in strArr {
let part = item.components(separatedBy: CharacterSet.decimalDigits.inverted).joined()
if let intVal = Int(part) {
print("this is a number -> \(intVal)")
}
}
Swift 4:
let string = "I have to buy 3 apples, 7 bananas, 10eggs"
let stringArray = string.components(separatedBy: CharacterSet.decimalDigits.inverted)
for item in stringArray {
if let number = Int(item) {
print("number: \(number)")
}
}
Using the "regex helper function" from Swift extract regex matches:
func matchesForRegexInText(regex: String!, text: String!) -> [String] {
let regex = NSRegularExpression(pattern: regex,
options: nil, error: nil)!
let nsString = text as NSString
let results = regex.matchesInString(text,
options: nil, range: NSMakeRange(0, nsString.length))
as! [NSTextCheckingResult]
return map(results) { nsString.substringWithRange($0.range)}
}
you can achieve that easily with
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let numbersAsStrings = matchesForRegexInText("\\d+", str) // [String]
let numbersAsInts = numbersAsStrings.map { $0.toInt()! } // [Int]
println(numbersAsInts) // [3, 7, 10]
The pattern "\d+" matches one or more decimal digit.
Of course the same can be done without the use of a helper function
if you prefer that for whatever reason:
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let regex = NSRegularExpression(pattern: "\\d+", options: nil, error: nil)!
let nsString = str as NSString
let results = regex.matchesInString(str, options: nil, range: NSMakeRange(0, nsString.length))
as! [NSTextCheckingResult]
let numbers = map(results) { nsString.substringWithRange($0.range).toInt()! }
println(numbers) // [3, 7, 10]
Alternative solution without regular expressions:
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let digits = "0123456789"
let numbers = split(str, allowEmptySlices: false) { !contains(digits, $0) }
.map { $0.toInt()! }
println(numbers) // [3, 7, 10]
let str = "Hello 1, World 62"
let intString = str.componentsSeparatedByCharactersInSet(
NSCharacterSet
.decimalDigitCharacterSet()
.invertedSet)
.joinWithSeparator("")
That will get you a string with all the number then you can just do this:
let int = Int(intString)
Just make sure you unwrap it since let int = Int(intString) is an optional.
For me makes more sense to have it as a String extension, probably it's a matter of tastes:
extension String {
func parseToInt() -> Int? {
return Int(self.components(separatedBy: CharacterSet.decimalDigits.inverted).joined())
}
}
So can be used like this:
if let number = "0kaksd020dk2kfj2123".parseToInt() {
// Do something with this number
}
Adapting from #flashadvanced's answer,
I found that the following is shorter and simpler for me.
let str = "I have to buy 3 apples, 7 bananas, 10eggs"
let component = str.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let list = component.filter({ $0 != "" }) // filter out all the empty strings in the component
print(list)
Tried in in the play ground and it works
Hope it helps :)
Swift 2.2
let strArr = str.characters.split{$0 == " "}.map(String.init)
for item in strArr {
let components = item.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let part = components.joinWithSeparator("")
if let intVal = Int(part) {
print("this is a number -> \(intVal)")
}
}
// This will only work with single digit numbers. Works with “10eggs” (no space between number and word
var str = "I have to buy 3 apples, 7 bananas, 10eggs"
var ints: [Int] = []
for char:Character in str {
if let int = "\(char)".toInt(){
ints.append(int)
}
}
The trick here is that you can check if a string is an integer (but you can’t check if a character is).
By looping though every character of the string, use string interpolation to create a string from the character and check if that string cas be casted as a integer.
If it can be, add it to the array.
// This will work with multi digit numbers. Does NOT work with “10 eggs” (has to have a space between number and word)
var str = "I have to buy 3 apples, 7 bananas, 10 eggs"
var ints: [Int] = []
var strArray = split(str) {$0 == " "}
for subString in strArray{
if let int = subString.toInt(){
ints.append(int)
}
}
Here we split the string at any space and create an array of every substring that is in the long string.
We again check every string to see if it is (or can be casted as) an integer.
Thanks for everyone who answered to my question.
I was looking for a block of code which uses only swift grammar, because I'm learning grammar only now..
I got an answer for my question.Maybe it is not an easier way to solve, but it uses only swift language.
var article = "I have to buy 3 apples, 7 bananas, 10 eggs"
var charArray = Array(article)
var unitValue = 0
var total = 0
for char in charArray.reverse() {
if let number = "\(char)".toInt() {
if unitValue==0 {
unitValue = 1
}
else {
unitValue *= 10
}
total += number*unitValue
}
else {
unitValue = 0
}
}
println("I bought \(total) apples.")
Swift 5:
extension String {
var allNumbers: [Int] {
let numbersInString = self.components(separatedBy: .decimalDigits.inverted).filter { !$0.isEmpty }
return numbersInString.compactMap { Int($0) }
}
}
You can get all numbers like
var str = "I have to buy 3 apples, 7 bananas, 10eggs"
// numbers = [3, 7, 10]
numbers = str.allNumbers
How could I replace nth character of a String with another one?
func replace(myString:String, index:Int, newCharac:Character) -> String {
// Write correct code here
return modifiedString
}
For example, replace("House", 2, "r") should be equal to "Horse".
Solutions that use NSString methods will fail for any strings with multi-byte Unicode characters. Here are two Swift-native ways to approach the problem:
You can use the fact that a String is a sequence of Character to convert the string to an array, modify it, and convert the array back:
func replace(myString: String, _ index: Int, _ newChar: Character) -> String {
var chars = Array(myString) // gets an array of characters
chars[index] = newChar
let modifiedString = String(chars)
return modifiedString
}
replace("House", 2, "r")
// Horse
Alternately, you can step through the string yourself:
func replace(myString: String, _ index: Int, _ newChar: Character) -> String {
var modifiedString = String()
for (i, char) in myString.characters.enumerate() {
modifiedString += String((i == index) ? newChar : char)
}
return modifiedString
}
Since these stay entirely within Swift, they're both Unicode-safe:
replace("🏠🏡🏠🏡🏠", 2, "🐴")
// 🏠🏡🐴🏡🏠
In Swift 4 it's much easier.
let newString = oldString.prefix(n) + char + oldString.dropFirst(n + 1)
This is an example:
let oldString = "Hello, playground"
let newString = oldString.prefix(4) + "0" + oldString.dropFirst(5)
where the result is
Hell0, playground
The type of newString is Substring. Both prefix and dropFirst return Substring. Substring is a slice of a string, in other words, substrings are fast because you don't need to allocate memory for the content of the string, but the same storage space as the original string is used.
I've found this solution.
var string = "Cars"
let index = string.index(string.startIndex, offsetBy: 2)
string.replaceSubrange(index...index, with: "t")
print(string)
// Cats
Please see NateCook answer for more details
func replace(myString: String, _ index: Int, _ newChar: Character) -> String {
var chars = Array(myString.characters) // gets an array of characters
chars[index] = newChar
let modifiedString = String(chars)
return modifiedString
}
For Swift 5
func replace(myString: String, _ index: Int, _ newChar: Character) -> String {
var chars = Array(myString) // gets an array of characters
chars[index] = newChar
let modifiedString = String(chars)
return modifiedString
}
replace("House", 2, "r")
This is no longer valid and deprecated.
You can always use swift String with NSString.So you can call NSString function on swift String.
By old stringByReplacingCharactersInRange: you can do like this
var st :String = "House"
let abc = st.bridgeToObjectiveC().stringByReplacingCharactersInRange(NSMakeRange(2,1), withString:"r") //Will give Horse
For modify existing string:
extension String {
subscript(_ n: Int) -> Character {
get {
let idx = self.index(startIndex, offsetBy: n)
return self[idx]
}
set {
let idx = self.index(startIndex, offsetBy: n)
self.replaceSubrange(idx...idx, with: [newValue])
}
}
}
var s = "12345"
print(s[0])
s[0] = "9"
print(s)
I've expanded upon Nate Cooks answer and transformed it into a string extension.
extension String {
//Enables replacement of the character at a specified position within a string
func replace(_ index: Int, _ newChar: Character) -> String {
var chars = Array(characters)
chars[index] = newChar
let modifiedString = String(chars)
return modifiedString
}
}
usage:
let source = "House"
let result = source.replace(2,"r")
result is "Horse"
I think what #Greg was trying to achieve with his extension is this:
mutating func replace(characterAt index: Int, with newChar: Character) {
var chars = Array(characters)
if index >= 0 && index < self.characters.count {
chars[index] = newChar
let modifiedString = String(chars)
self = modifiedString
} else {
print("can't replace character, its' index out of range!")
}
}
usage:
let source = "House"
source.replace(characterAt: 2, with: "r") //gives you "Horse"
After looking at the Swift Docs, I managed to make this function:
//Main function
func replace(myString:String, index:Int, newCharac:Character) -> String {
//Looping through the characters in myString
var i = 0
for character in myString {
//Checking to see if the index of the character is the one we're looking for
if i == index {
//Found it! Now instead of adding it, add newCharac!
modifiedString += newCharac
} else {
modifiedString += character
}
i = i + 1
}
// Write correct code here
return modifiedString
}
Please note that this is untested, but it should give you the right idea.
func replace(myString:String, index:Int, newCharac:Character) -> String {
var modifiedString = myString
let range = Range<String.Index>(
start: advance(myString.startIndex, index),
end: advance(myString.startIndex, index + 1))
modifiedString.replaceRange(range, with: "\(newCharac)")
return modifiedString
}
I would prefer to pass a String than a Character though.
Here's a way to replace a single character:
var string = "This is the original string."
let offset = 27
let index = string.index(string.startIndex, offsetBy: offset)
let range = index...index
print("ORIGINAL string: " + string)
string.replaceSubrange(range, with: "!")
print("UPDATED string: " + string)
// ORIGINAL string: This is the original string.
// UPDATED string: This is the original string!
This works with multi-character strings as well:
var string = "This is the original string."
let offset = 7
let index = string.index(string.startIndex, offsetBy: offset)
let range = index...index
print("ORIGINAL string: " + string)
string.replaceSubrange(range, with: " NOT ")
print("UPDATED string: " + string)
// ORIGINAL string: This is the original string.
// UPDATED string: This is NOT the original string.
var s = "helloworld"
let index = ((s.count) / 2) // index is 4
let firstIndex = s.index(s.startIndex, offsetBy: index)
let secondIndex = s.index(s.startIndex, offsetBy: index)
s.replaceSubrange(firstIndex...secondIndex, with: "*")
print("Replaced string is: \(s)") //OUTPUT IS: hell*world
This is working fine to replace string using the index.
String class in Swift (till v5 and maybe later) is what other languages call a StringBuilder class, and for performance reasons, Swift does NOT provide setting character by index; If you don't care about performance a simple solution could be:
public static func replace(_ string: String, at index: Int, with value: String) {
let start = string.index(string.startIndex, offsetBy: index)
let end = string.index(start, offsetBy: 1)
string.replaceSubrange(start..<end, with: value)
}
Or as an extension:
extension String {
public func charAt(_ index: Int) -> Character {
return self[self.index(self.startIndex, offsetBy: index)];
}
public mutating func setCharAt(_ index: Int, _ new: Character) {
self.setCharAt(index, String(new))
}
public mutating func setCharAt(_ index: Int, _ new: String) {
let i = self.index(self.startIndex, offsetBy: index)
self.replaceSubrange(i...i, with: new)
}
}
Note how above needs to call index(...) method to convert integer to actual-index!? It seems, Swift implements String like a linked-list, where append(...) is really fast, but even finding the index (without doing anything with it) is a linear-time operation (and gets slower based on concatenation count).
public void createEncodedSentence() {
StringBuffer buff = new StringBuffer();
int counter = 0;
char a;
for (int i = 0; i < sentence.length(); i++) {
a = sentence.charAt(i);
if (a == '.') {
buff.append('*');
}
if (a != ' ' && a != '.') {
counter++;
}
if (counter % 3 == 0) {
buff.append("");
}
buff.append(sentence.charAt(i));
}
encodedSentence = buff.toString();
}
Strings in swift don't have an accessor to read or write a single character. There's an excellent blog post by Ole Begemann describing how strings in swift work.
Note: the implementation below is wrong, read addendum
So the right way is by taking the left part of the string up to the index -1 character, append the replacing character, then append the string from index + 1 up to the end:
func myReplace(myString:String, index:Int, newCharac:Character) -> String {
var modifiedString: String
let len = countElements(myString)
if (index < len) && (index >= 0) {
modifiedString = myString.substringToIndex(index) + newCharac + myString.substringFromIndex(index + 1)
} else {
modifiedString = myString
}
return modifiedString
}
Note: in my implementation I chose to return the original string if the index is not in a valid range
Addendum Thanks to #slazyk, who found out that my implementation is wrong (see comment), I am providing a new swift only version of the function.
func replace(myString:String, index:Int, newCharac:Character) -> String {
var modifiedString: String
if (index < 0) || (index >= countElements(myString)) {
modifiedString = myString
} else {
var start = myString.startIndex
var end = advance(start, index)
modifiedString = myString[start ..< end]
modifiedString += newCharac
start = end.successor()
end = myString.endIndex
modifiedString += myString[start ... end]
}
return modifiedString
}
#codester's answer looks very good, and it's probably what I would use myself.
It would be interesting to know how performances compare though, using a fully swift solution and bridging to objective-c instead.
Here is an efficient answer :
import Foundation
func replace(myString:String, index:Int, newCharac:Character) -> String {
return myString.substringToIndex(index-1) + newCharac + myString.substringFromIndex(index)
}
I made a console program, but the problem is that it doesn't allow parameters to be inserted. So I'm wondering how would I split a single string into multiple strings to achieve what I need. E.g.: text="msg Hello" would be split into textA="msg" and textB="Hello"
This is the main console code so far (just to show the idea):
if (keyboard_check_pressed(vk_enter)) {
text_console_c = asset_get_index("scr_local_"+string(keyboard_string));
if (text_console_c > -1) {
text_console+= "> "+keyboard_string+"#";
script_execute(text_console_c);
text_console_c = -1;
}
else if (keyboard_string = "") {
text_console+= ">#";
}
else {
text_console+= "> Unknown command: "+keyboard_string+"#";
};
keyboard_string = "";
}
I cant recommend spliting string with iteration by char, because when u try split very very very long string, then time to split is very long and can freeze thread for a short/long time. Game maker is single threaded for now.
This code is much faster.
string_split
var str = argument[0] //string to split
var delimiter = argument[1] // delimiter
var letDelimiter = false // append delimiter to each part
if(argument_count == 3)
letDelimiter = argument[2]
var list = ds_list_create()
var d_at = string_pos(delimiter, str)
while(d_at > 0) {
var part = string_delete(str, d_at , string_length(str))
if(letDelimiter)
part = part + delimiter
str = string_delete(str, 1, d_at)
d_at = string_pos(delimiter, str)
ds_list_add(list, part)
if(d_at == 0 && str != "")//last string without delimiter, need to add too
ds_list_add(list, str)
}
return list;
Dont forget ds_list_destroy after you iterate all strings
for example:
var splited = string_split("first part|second part", '|')
for(splited) {
//do something with each string
}
ds_list_destroy(splited)
Something like this may help, haven't tested it out but if you can follow what is going on its a good place to start.
Text = "msg Hello"
counter = 0
stringIndex = 0
for (i = 0; i < string_length(text); i++)
{
if string_char_at(text,i) == " "
{
counter++
stringIndex = 0
} else {
string_insert(string_char_at(text,i),allStrings(counter),stringIndex)
stringIndex++
}
}
allStrings should be an array containing each of the separate strings. Whenever a " " is seen the next index of allStrings starts having it's characters filled in. stringIndex is used to add the progressive characters.
How can I remove last character from String variable using Swift? Can't find it in documentation.
Here is full example:
var expression = "45+22"
expression = expression.substringToIndex(countElements(expression) - 1)
Swift 4.0 (also Swift 5.0)
var str = "Hello, World" // "Hello, World"
str.dropLast() // "Hello, Worl" (non-modifying)
str // "Hello, World"
String(str.dropLast()) // "Hello, Worl"
str.remove(at: str.index(before: str.endIndex)) // "d"
str // "Hello, Worl" (modifying)
Swift 3.0
The APIs have gotten a bit more swifty, and as a result the Foundation extension has changed a bit:
var name: String = "Dolphin"
var truncated = name.substring(to: name.index(before: name.endIndex))
print(name) // "Dolphin"
print(truncated) // "Dolphi"
Or the in-place version:
var name: String = "Dolphin"
name.remove(at: name.index(before: name.endIndex))
print(name) // "Dolphi"
Thanks Zmey, Rob Allen!
Swift 2.0+ Way
There are a few ways to accomplish this:
Via the Foundation extension, despite not being part of the Swift library:
var name: String = "Dolphin"
var truncated = name.substringToIndex(name.endIndex.predecessor())
print(name) // "Dolphin"
print(truncated) // "Dolphi"
Using the removeRange() method (which alters the name):
var name: String = "Dolphin"
name.removeAtIndex(name.endIndex.predecessor())
print(name) // "Dolphi"
Using the dropLast() function:
var name: String = "Dolphin"
var truncated = String(name.characters.dropLast())
print(name) // "Dolphin"
print(truncated) // "Dolphi"
Old String.Index (Xcode 6 Beta 4 +) Way
Since String types in Swift aim to provide excellent UTF-8 support, you can no longer access character indexes/ranges/substrings using Int types. Instead, you use String.Index:
let name: String = "Dolphin"
let stringLength = count(name) // Since swift1.2 `countElements` became `count`
let substringIndex = stringLength - 1
name.substringToIndex(advance(name.startIndex, substringIndex)) // "Dolphi"
Alternatively (for a more practical, but less educational example) you can use endIndex:
let name: String = "Dolphin"
name.substringToIndex(name.endIndex.predecessor()) // "Dolphi"
Note: I found this to be a great starting point for understanding String.Index
Old (pre-Beta 4) Way
You can simply use the substringToIndex() function, providing it one less than the length of the String:
let name: String = "Dolphin"
name.substringToIndex(countElements(name) - 1) // "Dolphi"
The global dropLast() function works on sequences and therefore on Strings:
var expression = "45+22"
expression = dropLast(expression) // "45+2"
// in Swift 2.0 (according to cromanelli's comment below)
expression = String(expression.characters.dropLast())
Swift 4:
let choppedString = String(theString.dropLast())
In Swift 2, do this:
let choppedString = String(theString.characters.dropLast())
I recommend this link to get an understanding of Swift strings.
Swift 4/5
var str = "bla"
str.removeLast() // returns "a"; str is now "bl"
This is a String Extension Form:
extension String {
func removeCharsFromEnd(count_:Int) -> String {
let stringLength = count(self)
let substringIndex = (stringLength < count_) ? 0 : stringLength - count_
return self.substringToIndex(advance(self.startIndex, substringIndex))
}
}
for versions of Swift earlier than 1.2:
...
let stringLength = countElements(self)
...
Usage:
var str_1 = "Maxim"
println("output: \(str_1.removeCharsFromEnd(1))") // "Maxi"
println("output: \(str_1.removeCharsFromEnd(3))") // "Ma"
println("output: \(str_1.removeCharsFromEnd(8))") // ""
Reference:
Extensions add new functionality to an existing class, structure, or enumeration type. This includes the ability to extend types for which you do not have access to the original source code (known as retroactive modeling). Extensions are similar to categories in Objective-C. (Unlike Objective-C categories, Swift extensions do not have names.)
See DOCS
Use the function removeAtIndex(i: String.Index) -> Character:
var s = "abc"
s.removeAtIndex(s.endIndex.predecessor()) // "ab"
Swift 4
var welcome = "Hello World!"
welcome = String(welcome[..<welcome.index(before:welcome.endIndex)])
or
welcome.remove(at: welcome.index(before: welcome.endIndex))
or
welcome = String(welcome.dropLast())
The easiest way to trim the last character of the string is:
title = title[title.startIndex ..< title.endIndex.advancedBy(-1)]
import UIKit
var str1 = "Hello, playground"
str1.removeLast()
print(str1)
var str2 = "Hello, playground"
str2.removeLast(3)
print(str2)
var str3 = "Hello, playground"
str3.removeFirst(2)
print(str3)
Output:-
Hello, playgroun
Hello, playgro
llo, playground
let str = "abc"
let substr = str.substringToIndex(str.endIndex.predecessor()) // "ab"
var str = "Hello, playground"
extension String {
var stringByDeletingLastCharacter: String {
return dropLast(self)
}
}
println(str.stringByDeletingLastCharacter) // "Hello, playgroun"
Short answer (valid as of 2015-04-16): removeAtIndex(myString.endIndex.predecessor())
Example:
var howToBeHappy = "Practice compassion, attention and gratitude. And smile!!"
howToBeHappy.removeAtIndex(howToBeHappy.endIndex.predecessor())
println(howToBeHappy)
// "Practice compassion, attention and gratitude. And smile!"
Meta:
The language continues its rapid evolution, making the half-life for many formerly-good S.O. answers dangerously brief. It's always best to learn the language and refer to real documentation.
With the new Substring type usage:
Swift 4:
var before: String = "Hello world!"
var lastCharIndex: Int = before.endIndex
var after:String = String(before[..<lastCharIndex])
print(after) // Hello world
Shorter way:
var before: String = "Hello world!"
after = String(before[..<before.endIndex])
print(after) // Hello world
Use the function advance(startIndex, endIndex):
var str = "45+22"
str = str.substringToIndex(advance(str.startIndex, countElements(str) - 1))
A swift category that's mutating:
extension String {
mutating func removeCharsFromEnd(removeCount:Int)
{
let stringLength = count(self)
let substringIndex = max(0, stringLength - removeCount)
self = self.substringToIndex(advance(self.startIndex, substringIndex))
}
}
Use:
var myString = "abcd"
myString.removeCharsFromEnd(2)
println(myString) // "ab"
Another way If you want to remove one or more than one character from the end.
var myStr = "Hello World!"
myStr = (myStr as NSString).substringToIndex((myStr as NSString).length-XX)
Where XX is the number of characters you want to remove.
Swift 3 (according to the docs) 20th Nov 2016
let range = expression.index(expression.endIndex, offsetBy: -numberOfCharactersToRemove)..<expression.endIndex
expression.removeSubrange(range)
The dropLast() function removes the last element of the string.
var expression = "45+22"
expression = expression.dropLast()
Swift 4.2
I also delete my last character from String (i.e. UILabel text) in IOS app
#IBOutlet weak var labelText: UILabel! // Do Connection with UILabel
#IBAction func whenXButtonPress(_ sender: UIButton) { // Do Connection With X Button
labelText.text = String((labelText.text?.dropLast())!) // Delete the last caracter and assign it
}
I'd recommend using NSString for strings that you want to manipulate. Actually come to think of it as a developer I've never run into a problem with NSString that Swift String would solve... I understand the subtleties. But I've yet to have an actual need for them.
var foo = someSwiftString as NSString
or
var foo = "Foo" as NSString
or
var foo: NSString = "blah"
And then the whole world of simple NSString string operations is open to you.
As answer to the question
// check bounds before you do this, e.g. foo.length > 0
// Note shortFoo is of type NSString
var shortFoo = foo.substringToIndex(foo.length-1)
Swift 3: When you want to remove trailing string:
func replaceSuffix(_ suffix: String, replacement: String) -> String {
if hasSuffix(suffix) {
let sufsize = suffix.count < count ? -suffix.count : 0
let toIndex = index(endIndex, offsetBy: sufsize)
return substring(to: toIndex) + replacement
}
else
{
return self
}
}
complimentary to the above code I wanted to remove the beginning of the string and could not find a reference anywhere. Here is how I did it:
var mac = peripheral.identifier.description
let range = mac.startIndex..<mac.endIndex.advancedBy(-50)
mac.removeRange(range) // trim 17 characters from the beginning
let txPower = peripheral.advertisements.txPower?.description
This trims 17 characters from the beginning of the string (he total string length is 67 we advance -50 from the end and there you have it.
I prefer the below implementation because I don't have to worry even if the string is empty
let str = "abc"
str.popLast()
// Prints ab
str = ""
str.popLast() // It returns the Character? which is an optional
// Print <emptystring>