I'm trying to create a substring of the first 4 characters entered in a textField in Swift 4 on my iOS app.
Since the change to Swift 4 I'm struggling with basic String parsing.
So based on Apple documentation I'm assuming I need to use the substring.index function and I understand the second parameter (offsetBy) is the number of characters to create a substring with. I'm just unsure how I tell Swift to start at the beginning of the string.
This is the code so far:
let postcode = textFieldPostcode.text
let newPostcode = postcode?.index(STARTATTHEBEGININGOFTHESTRING, offsetBy: 4)
I hope my explanation makes sense, happy to answer any questions on this.
Thanks,
In Swift 4 you can use
let string = "Hello World"
let first4 = string.prefix(4) // Hell
The type of the result is a new type Substring which behaves very similar to String. However if first4 is supposed to leave the current scope โ for example as a return value of a function โ it's recommended to create a String explicitly:
let first4 = String(string.prefix(4)) // Hell
See also SE 0163 String Revision 1
In Swift 4:
let postcode = textFieldPostcode.text!
let index = postcode.index(postcode.startIndex, offsetBy: 4)
let newPostCode = String(postcode[..<index])
Related
I have a function that finds the current word a user has selected in a UITextView. However, if I call this function when an emoji is in the UITextView.text property, I see a crash. I believe this is because of the different character counts in String vs NSString.
How do I properly convert this?
func currentWord() -> String {
let cursorPosition = selectedRange.location
let separationCharacters = NSCharacterSet(charactersInString: " ")
// crash occurs here
let beginRange = Range(text.startIndex.advancedBy(0) ..< text.startIndex.advancedBy(cursorPosition))
let endRange = Range(text.startIndex.advancedBy(cursorPosition) ..< text.startIndex.advancedBy(text.characters.count))
let beginPhrase = text.substringWithRange(beginRange)
let endPhrase = text.substringWithRange(endRange)
let beginWords = beginPhrase.componentsSeparatedByCharactersInSet(separationCharacters)
let endWords = endPhrase.componentsSeparatedByCharactersInSet(separationCharacters)
return beginWords.last! + endWords.first!
}
I believe this is because of the different character counts in String vs NSString
You're right about that. You are shifting back and forth between using NSRange (Cocoa) and Range (Swift) โ and they work differently. And NSString (Cocoa) and String (Swift) have different ideas of where the character boundaries are. You need to be consistent.
Once you've used selectedRange in the first line, you are in the Cocoa world of NSRange. You need to stay consistently in the Cocoa world. Don't use any Swift Ranges! Don't use any Swift characters!
Form your beginRange entirely using NSRange โ for example, call NSMakeRange. Don't use characters.count; stay in the NSString world and use the string's length (in Swift, that is its utf16.count). Then all will be well.
I entered a two emojis in textfield ๐จโ๐จโ๐งโ๐ง๐, here I'm getting total number of 5 characters length whereas 4 characters for first emoji and 1 character for second. Looks like apple has combined 4 emojis to form a one.
I'm looking for the swift code where I can separate each of emojis separately, suppose by taking the above example I should be getting 2 strings/character separately for each emoji.
Can any one help me to solve this, I've tried many things like regex separation or componentsSeparatedByString or characterSet. but unfortunately ended up with negative.
Thanks in advance.
Update for Swift 4 (Xcode 9)
As of Swift 4 (tested with Xcode 9 beta) a "Emoji ZWJ Sequence" is
treated as a single Character as mandated by the Unicode 9 standard:
let str = "๐จโ๐จโ๐งโ๐ง๐"
print(str.count) // 2
print(Array(str)) // ["๐จโ๐จโ๐งโ๐ง", "๐"]
Also String is a collection of its characters (again), so we can
call str.count to get the length, and Array(str) to get all
characters as an array.
(Old answer for Swift 3 and earlier)
This is only a partial answer which may help in this particular case.
"๐จโ๐จโ๐งโ๐ง" is indeed a combination of four separate characters:
let str = "๐จโ๐จโ๐งโ๐ง๐" //
print(Array(str.characters))
// Output: ["๐จโ", "๐จโ", "๐งโ", "๐ง", "๐"]
which are glued together with U+200D (ZERO WIDTH JOINER):
for c in str.unicodeScalars {
print(String(c.value, radix: 16))
}
/* Output:
1f468
200d
1f468
200d
1f467
200d
1f467
1f60d
*/
Enumerating the string with the .ByComposedCharacterSequences
options combines these characters correctly:
var chars : [String] = []
str.enumerateSubstringsInRange(str.characters.indices, options: .ByComposedCharacterSequences) {
(substring, _, _, _) -> () in
chars.append(substring!)
}
print(chars)
// Output: ["๐จโ๐จโ๐งโ๐ง", "๐"]
But there are other cases where this does not work,
e.g. the "flags" which are a sequence of "Regional Indicator
characters" (compare Swift countElements() return incorrect value when count flag emoji). With
let str = "๐ฉ๐ช"
the result of the above loop is
["๐ฉ", "๐ช"]
which is not the desired result.
The full rules are defined in "3 Grapheme Cluster Boundaries"
in the "Standard Annex #29 UNICODE TEXT SEGMENTATION" in the
Unicode standard.
You can use this code example or this pod.
To use it in Swift, import the category into the YourProject_Bridging_Header
#import "NSString+EMOEmoji.h"
Then you can check the range for every emoji in your String:
let example: NSString = "๐จโ๐จโ๐งโ๐ง๐" // your string
let ranges: NSArray = example.emo_emojiRanges() // ranges of the emojis
for value in ranges {
let range:NSRange = (value as! NSValue).rangeValue
print(example.substringWithRange(range))
}
// Output: ["๐จโ๐จโ๐งโ๐ง", "๐"]
I created an small example project with the code above.
For further reading, this interesting article from Instagram.
I was able to get the phone number from the textfield with the method below in Swift 1.2 but I'm trying the same method in Swift 2 but I'm getting an error from .join "join is unavailable". Please how can I write the same method in Swift 2?
let userNumber = NSNumberFormatter().numberFromString("".join(phoneNumberTextField.text!.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)))
This answer didn't help me that I can't get join at all.
Cannot invoke `join` with an argument list of type (String, [String]) in Swift 2.0
Try this:
let numbers = phoneNumberTextField.text!.componentsSeparatedByCharactersInSet(NSCharacterSet.decimalDigitCharacterSet().invertedSet)
let userNumber = numbers.joinWithSeparator(" ") // Using space as separator
In a UITableView, I'm listing a bunch of languages to be selected. And to put a section index view to the right like in Contacts app, I'm getting all first letters of languages in the list and then use it to generate the section index view.
It works almost perfect, Just I encountered with a problem in getting first letter of some strings in Hebrew. Here a screenshot from playground, one of the language name that I couldn't get the first letter:
Problem is, the first letter of the name of the language that has "ina" language code, isn't "ื", it's an empty character; it's not a space, it's just an empty character. As you can see, it's actually 12 characters in total, but when I get count of it, it says 13 characters because there is an non-space empty character in index 0.
It works perfectly if I use "eng" or "ara" languages with putting these values in value: parameter. So maybe the problem is cause of system that returns a language name with an empty character in some cases, I don't know.
I tried some different methods of getting first letter, but any of it didn't work.
Here "ื" isn't the first letter, it's the second letter. So I thought maybe I can find a simple hack with that, but I want to try solving it before trying workarounds.
Here is the code:
let locale = NSLocale(localeIdentifier: "he")
let languageName = locale.displayNameForKey(NSLocaleIdentifier, value: "ina")!
let firstLetter = first(languageName)!
println(countElements(languageName))
for character in languageName {
println(character)
}
You could use an NSCharacterSet.controlCharacterSet() to test each character. I can't figure out how to stay in Swift-native strings, but here's a function that uses NSString to return the first non-control character:
func firstNonControlCharacter(str: NSString) -> String? {
let controlChars = NSCharacterSet.controlCharacterSet()
for i in 0..<str.length {
if !controlChars.characterIsMember(str.characterAtIndex(i)) {
return str.substringWithRange(NSRange(location: i, length: 1))
}
}
return nil
}
let locale = NSLocale(localeIdentifier: "he")
let languageName = locale.displayNameForKey(NSLocaleIdentifier, value: "ina")!
let firstChar = firstNonControlCharacter(languageName) // Optional("ื")
I'm playing around with emojis in Swift using Xcode playground for some simple iOS8 apps. For this, I want to create something similar to a unicode/emoji map/description.
In order to do this, I need to have a loop that would allow me to print out a list of emojis. I was thinking of something along these lines
for i in 0x1F601 - 0x1F64F {
var hex = String(format:"%2X", i)
println("\u{\(hex)}") //Is there another way to create UTF8 string corresponding to emoji
}
But the println() throws an error
Expected '}'in \u{...} escape sequence.
Is there a simple way to do this which I am missing?
I understand that not all entries will correspond to an emoji. Also, I'm able create a lookup table with reference from http://apps.timwhitlock.info/emoji/tables/unicode, but I would like a lazy/easy method of achieving the same.
You can loop over those hex values with a Range: 0x1F601...0x1F64F and then create the Strings using a UnicodeScalar:
for i in 0x1F601...0x1F64F {
guard let scalar = UnicodeScalar(i) else { continue }
let c = String(scalar)
print(c)
}
Outputs:
๐๐๐๐๐
๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐ก๐ข๐ฃ๐ค๐ฅ๐ฆ๐ง๐จ๐ฉ๐ช๐ซ๐ฌ๐ญ๐ฎ๐ฏ๐ฐ๐ฑ๐ฒ๐ณ๐ด๐ต๐ถ๐ท๐ธ๐น๐บ๐ป๐ผ๐ฝ๐พ๐ฟ๐๐๐๐๐๐
๐๐๐๐๐๐๐๐๐๐
If you want all the emoji, just add another loop over an array of ranges:
// NOTE: These ranges are still just a subset of all the emoji characters;
// they seem to be all over the place...
let emojiRanges = [
0x1F601...0x1F64F,
0x2702...0x27B0,
0x1F680...0x1F6C0,
0x1F170...0x1F251
]
for range in emojiRanges {
for i in range {
guard let scalar = UnicodeScalar(i) else { continue }
let c = String(scalar)
print(c)
}
}
For those asking, the full list of available emojis can be found here: https://www.unicode.org/emoji/charts/full-emoji-list.html
A parsable list of unicode sequences for all emojis can be found in the emoji-sequences.txt file under the directory for the version you're interested in here: http://unicode.org/Public/emoji/
As of 9/15/2021 the latest version of the emoji standard available on Apple devices is 13.1.