How do I get the last n-characters in a string?
I've tried using:
var string = 'Dart is fun';
var newString = string.substring(-5);
But that does not seem to be correct
var newString = string.substring(string.length - 5);
Create an extension:
extension E on String {
String lastChars(int n) => substring(length - n);
}
Usage:
var source = 'Hello World';
var output = source.lastChars(5); // 'World'
While #Alexandre Ardhuin is correct, it is important to note that if the string has fewer than n characters, an exception will be raised:
Uncaught Error: RangeError: Value not in range: -5
It would behoove you to check the length before running it that way
String newString(String oldString, int n) {
if (oldString.length >= n) {
return oldString.substring(oldString.length - n)
} else {
// return whatever you want
}
}
While you're at it, you might also consider ensuring that the given string is not null.
oldString ??= '';
If you like one-liners, another options would be:
String newString = oldString.padLeft(n).substring(max(oldString.length - n, 0)).trim()
If you expect it to always return a string with length of n, you could pad it with whatever default value you want (.padLeft(n, '0')), or just leave off the trim().
At least, as of Dart SDK 2.8.1, that is the case. I know they are working on improving null safety and this might change in the future.
var newString = string.substring((string.length - 5).clamp(0, string.length));
note: I am using clamp in order to avoid Value Range Error. By that you are also immune to negative n-characters if that is somehow calculated.
In fact I wonder that dart does not have such clamp implemented within the substring method.
If you want to be null aware, just use:
var newString = string?.substring((string.length - 5).clamp(0, string.length));
I wrote my own solution to get any no of last n digits from a string of unknown length, for example the 5th to the last digit from an n digit string,
String bin='408 408 408 408 408 1888';// this is the your string
// this function is to remove space from the string and then reverse the
string, then convert it to a list
List reversed=bin.replaceAll(" ","").split('').reversed.toList();
//and then get the 0 to 4th digit meaning if you want to get say 6th to last digit, just pass 0,6 here and so on. This second reverse function, return the string to its initial arrangement
var list = reversed.sublist(0,4).reversed.toList();
var concatenate = StringBuffer();
// this function is to convert the list back to string
list.forEach((item){
concatenate.write(item);
});
print(concatenate);// concatenate is the string you need
i wrote code to get character when user enter in text field and do math with them
this :
#IBOutlet weak internal var textMeli: UITextField!
var myChar = textMeli.text
var numb = [myChar[0]*3 , myChar[1]*7]
but one is wrong
textMeli.text is a String.
myChar is a String.
You can't access a Character from a String using bracket notation.
Take a look at the documentation for the String structure.
You'll see that you can access the string's characters through the characters property. This will return a collection of Characters. Initalize a new array with the collection and you can then use bracket notation.
let string = "Foo"
let character = Array(string.characters)[0]
character will be of type Character.
You'll then need to convert the Character to some sort of number type (Float, Int, Double, etc.) to use multiplication.
Type is important in programming. Make sure you are keeping track so you know what function and properties you can use.
Off the soap box. It looks like your trying to take a string and convert it into a number. I would skip the steps of using characters. Have two text fields, one to accept the first number (as a String) and the other to accept the second number (as a String). Use a number formatter to convert your string to a number. A number formatter will return you an NSNumber. Checking out the documentation and you'll see that you can "convert" the NSNumber to any number type you want. Then you can use multiplication.
Something like this:
let firstNumberTextField: UITextField!
let secondNumberTextField: UITextField!
let numberFormatter = NumberFormatter()
let firstNumber = numberFormatter.number(from: firstNumberTextField.text!)
let secondNumber = numberFormatter.number(from: secondNumberTextField.text!)
let firstInt = firstNumber.integerValue //or whatever type of number you need
let secondInt = secondNumber.integerValue
let product = firstInt * secondInt
Dealing with Swift strings is kind of tricky because of the way they deal with Unicode and "grapheme clusters". You can't index into String objects using array syntax like that.
Swift also doesn't treat characters as interchangeable with 8 bit ints like C does, so you can't do math on characters like you're trying to do. You have to take a String and cast it to an Int type.
You could create an extension to the String class that WOULD let you use integer subscripts of strings:
extension String {
subscript (index: Int) -> String {
let first = self.startIndex
let startIndex = self.index(first, offsetBy: index)
let nextIndex = self.index(first, offsetBy: index + 1)
return self[startIndex ..< nextIndex]
}
}
And then:
let inputString = textMeli.text
let firstVal = Int(inputString[0])
let secondVal = Int(inputString[2])
and
let result = firstVal * 3 + secondVal * 7
Note that the subscript extension above is inefficient and would be a bad way to do any sort of "heavy lifting" string parsing. Each use of square bracket indexing has as bad as O(n) performance, meaning that traversing an entire string would give nearly O(n^2) performance, which is very bad.
The code above also lacks range checking or error handling. It will crash if you pass it a subscript out of range.
Note that its very strange to take multiple characters as input, then do math on the individual characters as if they are separate values. This seems like really bad user interface.
Why don't you step back from the details and tell us what you are trying to do at a higher level?
Does swift by default has a random number generator that returns the same number only once?
For example it picks a number in range 1,2,3,4,5. Return that random number (lets say 3). In the next loop, it picks only from 1,2,4,5 and so on.
EDIT:
This is what I ended up using. It returns an array of integers from 0-5 and was tested in playground. Note that this is meant to be used when you are picking from a large set of Integers and not just 6.
func generateRandomArray() -> [Int]{
var randomImages: [Int] = [Int]()
var newRandomNumber = Int(arc4random_uniform(UInt32(6)))
while (randomImages.count < 6) {
if (randomImages.contains(newRandomNumber) == false){
randomImages.append(newRandomNumber)
}
newRandomNumber = Int(arc4random_uniform(UInt32(6)))
}
return randomImages
}
This kind of generator is not called a "random number" generator, but usually
a "Shuffle" or "Permute".
You have to tell it how many items there are first. Otherwise what you are proposing doesn't make any sense!
see answer here: How do I shuffle an array in Swift?
You can do this trick, Add the range of numbers from which you want to get the random result..to an NSMutableArray or NSMutableSet(to make sure there is only 1 of it).
Iterate through the array-
for(int i=0;i<myMutableArray .count;i++){
int randomIndex = arc4random_uniform(myMutableArray.count);
int myNumber = [[myMutableArray objectAtIndex:randomIndex]intValue]; // edited
NSLog(#"My random number-%i",myNumber);//this is your generated random number
[myMutableArray removeObjectAtIndex:randomIndex];
}
I guess this would do the tric, But if you do rmember NSMutableArray cannot take Primitive data type, just Objects like NSNumber.
EDIT
this line is not written and is justifying that why i am converting the NSNumber back to an interger because while adding the integer in the Array it has to be converted to a NSNumber|
[ myMutableArray addObject:[NSNumber numberWithInt:2]];
So in the for loop when i am getting the object from myMutableArray i am getting an NSNumber and to simplify it i casted that Object (NSNumber object) back to an integer.
From Apple's documentation:
A set stores distinct values of the same type in a collection with no
defined ordering.
This seems to be exactly what you are after, so making an array seems a bit ill-advised...
var values : Set<Int> = []
var value : Int {
return Int(arc4random() % 6 + 1)
}
while values.count < 6 {
values.insert(value)
}
And you probably want to have a look at those values somewhere down the line:
for value in values {
println(value)
}
Just store the last random number in a variable and check the next random number against it in a while loop. While the new random number is the same as the last random number, generate a new random number. Once the new random number generated is not the same as the last random number the while loop ends and we update the last random number.
var lastRandomNumber = 0
var newRandomNumber = 0
func gernerateRandomNumber() {
newRandomNumber = Int(arc4random_uniform(5)+1)
while newRandomNumber == lastRandomNumber {
newRandomNumber = Int(arc4random_uniform(5)+1)
}
lastRandomNumber = newRandomNumber
}
I would like to create a function that looks at a string, and if it's a decimal string, returns it as a currency-formatted string. The function below does that, however if I pass in a string that is already formatted, it will fail of course (it expects to see a string like '25' or '25.55' but not '$15.25'
Is there a way to modify my function below to add another if condition that says "if you've already been formatted as a currency string, or your string is not in the right format, return X" (maybe X will be 0, or maybe it will be self (the same string) i'm not sure yet).
func toCurrencyStringFromDecimalString() -> String
{
var numberFormatter = NSNumberFormatter()
numberFormatter.numberStyle = NSNumberFormatterStyle.CurrencyStyle
if (self.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceCharacterSet()).utf16Count == 0)
{
//If whitespace is passed in, just return 0.0 as default
return numberFormatter.stringFromNumber(NSDecimalNumber(string: "0.0"))!
}
else if (IS_NOT_A_DECIMAL_OR_ALREADY_A_CURRENCY_STRING)
{
//So obviously this would go here to see if it's not a decimal (or already contains a current placeholder etc)
}
else
{
return numberFormatter.stringFromNumber(NSDecimalNumber(string: self))!
}
}
Thank you for your help!
Sounds like you need to use NSScanner.
According to the docs, the scanDecimal function of NSScanner:
Skips past excess digits in the case of overflow, so the receiver’s
position is past the entire integer representation.
Invoke this method with NULL as value to simply scan past a decimal integer representation.
I've been mostly programming in Obj-C so my Swift is rubbish, but here's my attempt at translating the appropriate code for detecting numeric strings (as also demonstrated in this answer):
let scanner: NSScanner = NSScanner(string:self)
let isNumeric = scanner.scanDecimal(nil) && scanner.atEnd
If the string is not a decimal representation, isNumeric should return false.
How do you get the length of a String? For example, I have a variable defined like:
var test1: String = "Scott"
However, I can't seem to find a length method on the string.
As of Swift 4+
It's just:
test1.count
for reasons.
(Thanks to Martin R)
As of Swift 2:
With Swift 2, Apple has changed global functions to protocol extensions, extensions that match any type conforming to a protocol. Thus the new syntax is:
test1.characters.count
(Thanks to JohnDifool for the heads up)
As of Swift 1
Use the count characters method:
let unusualMenagerie = "Koala 🐨, Snail 🐌, Penguin 🐧, Dromedary 🐪"
println("unusualMenagerie has \(count(unusualMenagerie)) characters")
// prints "unusualMenagerie has 40 characters"
right from the Apple Swift Guide
(note, for versions of Swift earlier than 1.2, this would be countElements(unusualMenagerie) instead)
for your variable, it would be
length = count(test1) // was countElements in earlier versions of Swift
Or you can use test1.utf16count
TLDR:
For Swift 2.0 and 3.0, use test1.characters.count. But, there are a few things you should know. So, read on.
Counting characters in Swift
Before Swift 2.0, count was a global function. As of Swift 2.0, it can be called as a member function.
test1.characters.count
It will return the actual number of Unicode characters in a String, so it's the most correct alternative in the sense that, if you'd print the string and count characters by hand, you'd get the same result.
However, because of the way Strings are implemented in Swift, characters don't always take up the same amount of memory, so be aware that this behaves quite differently than the usual character count methods in other languages.
For example, you can also use test1.utf16.count
But, as noted below, the returned value is not guaranteed to be the same as that of calling count on characters.
From the language reference:
Extended grapheme clusters can be composed of one or more Unicode
scalars. This means that different characters—and different
representations of the same character—can require different amounts of
memory to store. Because of this, characters in Swift do not each take
up the same amount of memory within a string’s representation. As a
result, the number of characters in a string cannot be calculated
without iterating through the string to determine its extended
grapheme cluster boundaries. If you are working with particularly long
string values, be aware that the characters property must iterate over
the Unicode scalars in the entire string in order to determine the
characters for that string.
The count of the characters returned by the characters property is not
always the same as the length property of an NSString that contains
the same characters. The length of an NSString is based on the number
of 16-bit code units within the string’s UTF-16 representation and not
the number of Unicode extended grapheme clusters within the string.
An example that perfectly illustrates the situation described above is that of checking the length of a string containing a single emoji character, as pointed out by n00neimp0rtant in the comments.
var emoji = "👍"
emoji.characters.count //returns 1
emoji.utf16.count //returns 2
Swift 1.2 Update: There's no longer a countElements for counting the size of collections. Just use the count function as a replacement: count("Swift")
Swift 2.0, 3.0 and 3.1:
let strLength = string.characters.count
Swift 4.2 (4.0 onwards): [Apple Documentation - Strings]
let strLength = string.count
Swift 1.1
extension String {
var length: Int { return countElements(self) } //
}
Swift 1.2
extension String {
var length: Int { return count(self) } //
}
Swift 2.0
extension String {
var length: Int { return characters.count } //
}
Swift 4.2
extension String {
var length: Int { return self.count }
}
let str = "Hello"
let count = str.length // returns 5 (Int)
Swift 4
"string".count
;)
Swift 3
extension String {
var length: Int {
return self.characters.count
}
}
usage
"string".length
If you are just trying to see if a string is empty or not (checking for length of 0), Swift offers a simple boolean test method on String
myString.isEmpty
The other side of this coin was people asking in ObjectiveC how to ask if a string was empty where the answer was to check for a length of 0:
NSString is empty
Swift 5.1, 5
let flag = "🇵🇷"
print(flag.count)
// Prints "1" -- Counts the characters and emoji as length 1
print(flag.unicodeScalars.count)
// Prints "2" -- Counts the unicode lenght ex. "A" is 65
print(flag.utf16.count)
// Prints "4"
print(flag.utf8.count)
// Prints "8"
tl;dr If you want the length of a String type in terms of the number of human-readable characters, use countElements(). If you want to know the length in terms of the number of extended grapheme clusters, use endIndex. Read on for details.
The String type is implemented as an ordered collection (i.e., sequence) of Unicode characters, and it conforms to the CollectionType protocol, which conforms to the _CollectionType protocol, which is the input type expected by countElements(). Therefore, countElements() can be called, passing a String type, and it will return the count of characters.
However, in conforming to CollectionType, which in turn conforms to _CollectionType, String also implements the startIndex and endIndex computed properties, which actually represent the position of the index before the first character cluster, and position of the index after the last character cluster, respectively. So, in the string "ABC", the position of the index before A is 0 and after C is 3. Therefore, endIndex = 3, which is also the length of the string.
So, endIndex can be used to get the length of any String type, then, right?
Well, not always...Unicode characters are actually extended grapheme clusters, which are sequences of one or more Unicode scalars combined to create a single human-readable character.
let circledStar: Character = "\u{2606}\u{20DD}" // ☆⃝
circledStar is a single character made up of U+2606 (a white star), and U+20DD (a combining enclosing circle). Let's create a String from circledStar and compare the results of countElements() and endIndex.
let circledStarString = "\(circledStar)"
countElements(circledStarString) // 1
circledStarString.endIndex // 2
In Swift 2.0 count doesn't work anymore. You can use this instead:
var testString = "Scott"
var length = testString.characters.count
Here's something shorter, and more natural than using a global function:
aString.utf16count
I don't know if it's available in beta 1, though. But it's definitely there in beta 2.
Updated for Xcode 6 beta 4, change method utf16count --> utf16Count
var test1: String = "Scott"
var length = test1.utf16Count
Or
var test1: String = "Scott"
var length = test1.lengthOfBytesUsingEncoding(NSUTF16StringEncoding)
As of Swift 1.2 utf16Count has been removed. You should now use the global count() function and pass the UTF16 view of the string. Example below...
let string = "Some string"
count(string.utf16)
For Xcode 7.3 and Swift 2.2.
let str = "🐶"
If you want the number of visual characters:
str.characters.count
If you want the "16-bit code units within the string’s UTF-16 representation":
str.utf16.count
Most of the time, 1 is what you need.
When would you need 2? I've found a use case for 2:
let regex = try! NSRegularExpression(pattern:"🐶",
options: NSRegularExpressionOptions.UseUnixLineSeparators)
let str = "🐶🐶🐶🐶🐶🐶"
let result = regex.stringByReplacingMatchesInString(str,
options: NSMatchingOptions.WithTransparentBounds,
range: NSMakeRange(0, str.utf16.count), withTemplate: "dog")
print(result) // dogdogdogdogdogdog
If you use 1, the result is incorrect:
let result = regex.stringByReplacingMatchesInString(str,
options: NSMatchingOptions.WithTransparentBounds,
range: NSMakeRange(0, str.characters.count), withTemplate: "dog")
print(result) // dogdogdog🐶🐶🐶
You could try like this
var test1: String = "Scott"
var length = test1.bridgeToObjectiveC().length
in Swift 2.x the following is how to find the length of a string
let findLength = "This is a string of text"
findLength.characters.count
returns 24
Swift 2.0:
Get a count: yourString.text.characters.count
Fun example of how this is useful would be to show a character countdown from some number (150 for example) in a UITextView:
func textViewDidChange(textView: UITextView) {
yourStringLabel.text = String(150 - yourStringTextView.text.characters.count)
}
In swift4 I have always used string.count till today I have found that
string.endIndex.encodedOffset
is the better substitution because it is faster - for 50 000 characters string is about 6 time faster than .count. The .count depends on the string length but .endIndex.encodedOffset doesn't.
But there is one NO. It is not good for strings with emojis, it will give wrong result, so only .count is correct.
In Swift 4 :
If the string does not contain unicode characters then use the following
let str : String = "abcd"
let count = str.count // output 4
If the string contains unicode chars then use the following :
let spain = "España"
let count1 = spain.count // output 6
let count2 = spain.utf8.count // output 7
In Xcode 6.1.1
extension String {
var length : Int { return self.utf16Count }
}
I think that brainiacs will change this on every minor version.
Get string value from your textview or textfield:
let textlengthstring = (yourtextview?.text)! as String
Find the count of the characters in the string:
let numberOfChars = textlength.characters.count
Here is what I ended up doing
let replacementTextAsDecimal = Double(string)
if string.characters.count > 0 &&
replacementTextAsDecimal == nil &&
replacementTextHasDecimalSeparator == nil {
return false
}
Swift 4 update comparing with swift 3
Swift 4 removes the need for a characters array on String. This means that you can directly call count on a string without getting characters array first.
"hello".count // 5
Whereas in swift 3, you will have to get characters array and then count element in that array. Note that this following method is still available in swift 4.0 as you can still call characters to access characters array of the given string
"hello".characters.count // 5
Swift 4.0 also adopts Unicode 9 and it can now interprets grapheme clusters. For example, counting on an emoji will give you 1 while in swift 3.0, you may get counts greater than 1.
"👍🏽".count // Swift 4.0 prints 1, Swift 3.0 prints 2
"👨❤️💋👨".count // Swift 4.0 prints 1, Swift 3.0 prints 4
Swift 4
let str = "Your name"
str.count
Remember: Space is also counted in the number
You can get the length simply by writing an extension:
extension String {
// MARK: Use if it's Swift 2
func stringLength(str: String) -> Int {
return str.characters.count
}
// MARK: Use if it's Swift 3
func stringLength(_ str: String) -> Int {
return str.characters.count
}
// MARK: Use if it's Swift 4
func stringLength(_ str: String) -> Int {
return str.count
}
}
Best way to count String in Swift is this:
var str = "Hello World"
var length = count(str.utf16)
String and NSString are toll free bridge so you can use all methods available to NSString with swift String
let x = "test" as NSString
let y : NSString = "string 2"
let lenx = x.count
let leny = y.count
test1.characters.count
will get you the number of letters/numbers etc in your string.
ex:
test1 = "StackOverflow"
print(test1.characters.count)
(prints "13")
Apple made it different from other major language. The current way is to call:
test1.characters.count
However, to be careful, when you say length you mean the count of characters not the count of bytes, because those two can be different when you use non-ascii characters.
For example;
"你好啊hi".characters.count will give you 5 but this is not the count of the bytes.
To get the real count of bytes, you need to do "你好啊hi".lengthOfBytes(using: String.Encoding.utf8). This will give you 11.
Right now (in Swift 2.3) if you use:
myString.characters.count
the method will return a "Distance" type, if you need the method to return an Integer you should type cast like so:
var count = myString.characters.count as Int
my two cents for swift 3/4
If You need to conditionally compile
#if swift(>=4.0)
let len = text.count
#else
let len = text.characters.count
#endif