Impossible to find the solution ; it does not work...
I've been on this for hours... A little help will give me the opportunity to sleep without a nightmare...
Where is the error ?
let num: Int = 128150 // smiley = "\u{1F496}" => 128150
var str: String = String(num, radix: 16)
str = str.uppercased()
var wkHex: String = "\\u{"+str+"}" // wkHex = "\u{"+str+"}" not match
wkHex.characters.removeFirst(0) // remove "\" unnecessary at startIndex
let cnt = wkHex.characters.count
let zzz: Array = Array(wkHex.characters)
var car: String = ""
for i in 0...cnt - 1 {
car.append(zzz[i])
}
outputChar.stringValue = car // outputChar is a Label (NSTextField)
// output : \u{1F496} ! instead of : 💖
So the idea is to go from a code point to a character?
let iii = 128150
let char = Character(UnicodeScalar(iii)!)
print(char) // 💖
Swift only allows you to use the \u{...} syntax at compile time. This means that the string won't be turned into the emoji at runtime, when the value of num is known.
To do this, you can use UnicodeScalar:
let unicode = UnicodeScalar(128150)
unicode?.description // 💖
Related
I would like to know if there is an efficient way of splitting a string into multiple strings based on delimiters that are also strings.
Eg. updateStr = "23+45 = 56 0" , delimiters = ["+"," ","="]
Result = [23,45,56,0]
I tried the following code in swift:
for i in 0..<delimiter.count {
let res = updateStr.components(separatedBy: delimiter[i])
updateStr = res.joined(separator: "unique%")
}
splitTxt = updateStr.components(separatedBy: "unique%")
This works, but as the delimiters will be received dynamically I want a better approach.
Are there any efficient ways to avoid multiple loops to solve this?
An algorithm with more efficient solution that doesn't involve swift instance methods would also be appreciated.
Thanks for the answers but
To be clearer, I don't just want characters but strings as delimiters:
Eg2. updateStr = "I like playing with friends" , delimiters = ["li"," "la","ie"]
Result = ["I ","ke p","ying with fr","nds"]
The efficient way to do this sort of thing is with a Set:
let equation = "23+45 = 56 0"
let delimiters : [Character] = ["+"," ","="]
let setOfSeparators = Set(delimiters)
let result = equation.split {setOfSeparators.contains($0)}
print(result)
That's efficient because contains on a Set is extremely fast, so that cost is negligible and we are looping implicitly through the original string just once.
On the other hand, you could take advantage of the Cocoa CharacterSet class. For that, I would say:
let equation = "23+45 = 56 0"
let delimiters = ["+"," ","="]
let characterSet = CharacterSet(charactersIn: delimiters.joined())
let result = equation.components(separatedBy: characterSet).filter {!$0.isEmpty}
print(result)
Another fun way is to use a Scanner (these are underutilized in my opinion):
let equation = "23+45 = 56 0"
let delimiters = ["+"," ","="]
let characterSet = CharacterSet(charactersIn: delimiters.joined())
let scanner = Scanner(string: equation)
var result = [String]()
while let word = scanner.scanUpToCharacters(from: characterSet) {
result.append(word)
scanner.scanCharacters(from: characterSet)
}
print(result)
One of the components(separatedBy:) overloads will handle this automatically using a CharacterSet:
let delimiters = ["+"," ","="].compactMap(UnicodeScalar.init)
let splitTxt = updateStr.components(separatedBy: CharacterSet(delimiters))
NSRegularExpression provides the facility to split on general regular expressions, so this would enable splitting on a finite set of string delimiters using a delim1|delim2|delim3 regex. The following split operation does this job:
static func stringSubrange(str : String, st : Int, en : Int) -> String
{ var result : [Character] = [Character]()
var count : Int = 0
for index in str.indices
{ let c : Character = str[index]
count = count + 1
if count >= st && count <= en
{ result.append(c) }
else if count > en
{ return String(result) }
}
return String(result)
}
static func split(str: String, pattern: String) -> [String]
{ let rge = NSRange(location: 0, length: str.utf16.count)
let regexp = try! NSRegularExpression(pattern: pattern)
let pred = regexp.matches(in: str, options: [], range: rge)
var result : [String] = [String]()
var prev : Int = 1;
for p in pred
{ let range = p.range
let splitString = Ocl.stringSubrange(str: str, st: prev, en: range.location)
prev = range.location + range.length + 1
if splitString.count > 0
{ result.append(splitString) }
}
if prev < str.count
{ result.append(Ocl.stringSubrange(str: str, st: prev, en: str.count)) }
return result
}
Last night I had to convert my Swift 2.3 code to Swift 3.0 and my code is a mess after the conversion.
In Swift 2.3 I had the following code:
let maxChar = 40;
let val = "some long string";
var startRange = val.startIndex;
var endRange = val.startIndex.advancedBy(maxChar, limit: val.endIndex);
let index = val.rangeOfString(" ", options: NSStringCompareOptions.BackwardsSearch , range: startRange...endRange , locale: nil)?.startIndex;
Xcode converted my code to this which doesn't work:
let maxChar = 40;
let val = "some long string";
var startRange = val.startIndex;
var endRange = val.characters.index(val.startIndex, offsetBy: maxChar, limitedBy: val.endIndex);
let index = val.range(of: " ", options: NSString.CompareOptions.backwards , range: startRange...endRange , locale: nil)?.lowerBound
The error is in the parameter range in val.rage, saying No '...' candidates produce the expected contextual result type 'Range?'.
I tried using Range(startRange...endRange) as suggestd in the docs but I'm getting en error saying: connot invoke initiliazer for type ClosedRange<_> with an arguement list of type (ClosedRange). Seems like I'm missing something fundametnal.
Any help is appreciated.
Thanks!
Simple answer: the fundamental thing you are missing is that a closed range is now different from a range. So, change startRange...endRange to startRange..<endRange.
In more detail, here's an abbreviated version of your code (without the maxChar part):
let val = "some long string";
var startRange = val.startIndex;
var endRange = val.endIndex;
let index = val.range(
of: " ", options: .backwards, range: startRange..<endRange)?.lowerBound
// 9
Now you can use that as a basis to restore your actual desired functionality.
However, if all you want to do is split the string, then reinventing the wheel is kind of silly:
let arr = "hey ho ha".characters.split(separator:" ").map{String($0)}
arr // ["hey", "ho", "ha"]
I'm getting a value like this 264.8 and I want to take the value before and after the dot. I can take the value before the dot like this
var string = "264.8"
var index = string2.rangeOfString(".", options: .BackwardsSearch)?.startIndex
var substring = string.substringToIndex(index2!)
but please how I can take it after the dot?
Try this code:
var string = "264.8"
var numbers = string.componentsSeparatedByString(".")
print(numbers[0])
print(numbers[1])
var string = "264.8"
let partsArr = string.componentsSeparatedByString(".")
var beforeDot: String = partsArr[0]
var afterDot: String? = partsArr[1]
Just for the sake of completeness, an alternative is to use split:
let string = "264.8"
let result = string.characters.split(".").map { String($0) }
print(result[0]) // "264"
print(result[1]) // "8"
And another one is to use componentsSeparatedByCharactersInSet:
let string = "264.8"
let result = string.componentsSeparatedByCharactersInSet(NSCharacterSet.punctuationCharacterSet())
print(result[0]) // "264"
print(result[1]) // "8"
Alternatively, define a closure variable that handles the conversion for you
let mySubstringClosure : (String) -> (String) = { $0.componentsSeparatedByString(".").first ?? $0 }
let substring1 = mySubstringClosure("264.8") // "264"
let substring2 = mySubstringClosure("264") // "264"
let substring3 = mySubstringClosure("") // ""
Note that this code runs safely even if no dot . exists in the string, or of the string is empty.
I am using the following code to implement basic dictionary using swift. However the compiler is not returning any values. I don't know what seems to be the problem. Need Help!
P.S I'm new to Swift.
import Foundation
var dic = ["Nil":"Neel Goswami","Kirana":"Kinara Shah","Sapre":"Rohan Sapre","JP":"Joy Patel","Shreya":"Shrey Bhat","Ali Bhai":"Aalisha Sheth","Gandhi":"Shlok Gandhi","Binti":"Biyanta Shah","Udgam":"Aayushi Shah"]
dic["Wary"] = "Aishwary Rawat"
dic["Sixer"] = "Ruchir Patel"
dic["Bhikhari"] = "Aabhas Singhal"
var str: String? = "Initial"
println("Enter the pet name: ")
str = NSString(data: NSFileHandle.fileHandleWithStandardInput().availableData, encoding:NSUTF8StringEncoding)
var st: String = str!
for (pet, act) in dic
{
if (pet == st) {
println("His/Her actual name is \(act)")
}
}
The problem is that the string from the user input contains a trailing newline character
(\n). You can fix that by changing
var st: String = str!
to
var st = str!.stringByTrimmingCharactersInSet(NSCharacterSet.newlineCharacterSet())
Alternatively, use
var st = str!.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet())
which removes leading and trailing space characters as well.
Note that you could simplify your for-loop to a dictionary lookup:
if let act = dic[st] {
println("His/Her actual name is \(act)")
}
This used to work in Xcode 6: Beta 5. Now I'm getting a compilation error in Beta 6.
for aCharacter: Character in aString {
var str: String = ""
var newStr: String = str.append(aCharacter) // ERROR
...
}
Error: Cannot invoke append with an argument of type Character
Update for the moving target that is Swift:
Swift no longer has a + operator that can take a String and an array of characters. (There is a string method appendContentsOf() that can be used for this purpose).
The best way of doing this now is Martin R’s answer in a comment below:
var newStr:String = str + String(aCharacter)
Original answer:
This changed in Beta 6. Check the release notes.I'm still downloading it, but try using:
var newStr:String = str + [aCharacter]
This also works
var newStr:String = str + String(aCharacter)
append append(c: Character) IS the right method but your code has two other problems.
The first is that to iterate over the characters of a string you must access the String.characters property.
The second is that the append method doesn't return anything so you should remove the newStr.
The code then looks like this:
for aCharacter : Character in aString.characters {
var str:String = ""
str.append(aCharacter)
// ... do other stuff
}
Another possible option is
var s: String = ""
var c: Character = "c"
s += "\(c)"
According to Swift 4 Documentation ,
You can append a Character value to a String variable with the String type’s append() method:
var welcome = "hello there"
let exclamationMark: Character = "!"
welcome.append(exclamationMark)
// welcome now equals "hello there!"
var stringName: String = "samontro"
var characterNameLast: Character = "n"
stringName += String(characterNameLast) // You get your name "samontron"
I had to get initials from first and last names, and join them together. Using bits and pieces of the above answers, this worked for me:
var initial: String = ""
if !givenName.isEmpty {
let char = (givenName as NSString).substring(with: NSMakeRange(0, 1))
let str = String(char)
initial.append(str)
}
if !familyName.isEmpty {
let char = (familyName as NSString).substring(with: NSMakeRange(0, 1))
let str = String(char)
initial.append(str)
}
for those looking for swift 5, you can do interpolation.
var content = "some random string"
content = "\(content)!!"
print(content) // Output: some random string!!
let original:String = "Hello"
var firstCha = original[original.startIndex...original.startIndex]
var str = "123456789"
let x = (str as NSString).substringWithRange(NSMakeRange(0, 4))
var appendString1 = "\(firstCha)\(x)" as String!
// final name
var namestr = "yogesh"
var appendString2 = "\(namestr) (\(appendString1))" as String!*