Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 7 years ago.
Improve this question
I have 2 enums below.
CONFIG_URLS.BASE_URL
CONFIG_URLS.URL1
Each of these enums points to a string.
I want to create a string variable by concatenating BASE_URL and URL1.
Should be swift code.
Can anybody help? Many thanks
I presume that your enum has a raw value of type string. In that case, I would recommend creating a static function that given a enum case returns an absolute URL obtained by appending the value of a case representing a path to the value of the base url:
enum CONFIG_URLS : String {
case BASE_URL = "http://www.myurl.com"
case URL1 = "/some/path"
static func getUrl(url: CONFIG_URLS) -> String {
switch url {
case .BASE_URL:
return BASE_URL.rawValue
default:
return "\(CONFIG_URLS.BASE_URL.rawValue)\(url.rawValue)"
}
}
}
println(CONFIG_URLS.getUrl(.BASE_URL))
println(CONFIG_URLS.getUrl(.URL1))
Alternatively, the static method can be converted to a property, used in a similar way:
enum CONFIG_URLS : String {
case BASE_URL = "http://www.myurl.com"
case URL1 = "/some/path"
var url: String {
switch self {
case .BASE_URL:
return BASE_URL.rawValue
default:
return "\(CONFIG_URLS.BASE_URL.rawValue)\(self.rawValue)"
}
}
}
println(CONFIG_URLS.BASE_URL.url)
println(CONFIG_URLS.URL1.url)
Swift provides the following options.
Option 1
You can concatenate two constant strings as below:
let str1 = "hi" // constant
let str2 = " how are u" // constant
var str3 = string1 + string2 // here str3 is variable which will hold
// the concatenated value i.e. str3 = "hi how are u"
Option 2
You can also append a String value to an existing String variable with the addition assignment operator (+=):
let str1 = "hi"
var str2 = "how are u"
str2 += str1
// str2 now equals "hi how are u"
THe solution was to use rawValue and use + for concatenation (in Swift) , which I wasnt aware about regarding enums.
Thanks everyone.
Related
This question already has answers here:
How to capitalize each word in a string using Swift iOS
(8 answers)
Closed 5 years ago.
How can i capitalized each first letter of the result of this
self.namE.text = currentUser.displayName
self.handle.text = snapshotValue?["handle"] as? String
instead of "Bruce willis" i would like to have "Bruce Willis", i created this extension to capitalized the first letter
extension String {
func capitalizingFirstLetter() -> String {
return prefix(1).uppercased() + dropFirst()
}
}
but it obviously capitalize only the first word in a string, so how i have to modify this extension to get the right result? ( i looked in the doc but i didn't understand very well )
String in swift4 has already a capitalized computed property on itself, so without implementing anything yourself, you can get the desired result using this:
self.namE.text = currentUser.displayName.capitalized
E.g.:
self.namE.text = "bruce willis".capitalized
characters - an instance property of String, is deprecated from with Xcode 9.1
It was very useful to get a substring from String by using the characters property but now it has been deprecated and Xcode suggests to use substring. I've tried to check around SO questions and apple developer tutorials/guidelines for the same. But could not see any solution/alternate as suggested.
Here is warning message:
'characters' is deprecated: Please use String or Substring
I've so many string operations are performed/handled using property characters.
Anyone have any idea/info about this update?
Swift 4 introduced changes on string API.
You can just use !stringValue.isEmpty instead of stringValue.characters.count > 0
for more information you get the sample from here
for e.g
let edit = "Summary"
edit.count // 7
Swift 4 vs Swift 3 examples:
let myString = "test"
for char in myString.characters {print(char) } // Swift 3
for char in myString { print(char) } // Swift 4
let length = myString.characters.count // Swift 3
let length = myString.count // Swift 4
One of the most common cases for manipulating strings is with JSON responses. In this example I created an extension in my watch app to drop the last (n) characters of a Bitcoin JSON object.
Swift 3:
func dropLast(_ n: Int = 0) -> String {
return String(characters.dropLast(n))
Xcode 9.1 Error Message:
'characters' is deprecated: Please use String or Substring directly
Xcode is telling us to use the string variable or method directly.
Swift 4:
func dropLast(_ n: Int = 0) -> String {
return String(dropLast(n))
}
Complete Extension:
extension String {
func dropLast(_ n: Int = 0) -> String {
return String(dropLast(n))
}
var dropLast: String {
return dropLast()
}
}
Call:
print("rate:\(response.USDRate)")
let literalMarketPrice = response.USDRate.dropLast(2)
print("literal market price: \(literalMarketPrice)")
Console:
//rate:7,101.0888 //JSON float
//literal market price: 7,101.08 // JSON string literal
Additional Examples:
print("Spell has \(invisibleSpellName.count) characters.")
return String(dropLast(n))
return String(removeLast(n))
Documentation:
You'll often be using common methods such as dropLast() or removeLast() or count so here is the explicit Apple documentation for each method.
droplast()
removelast()
counting characters
Use this characters because String stopped being a collection in Swift 2.0. However this is still valid code in Swift 4 but is no longer necessary now that String is a Collection again.
For example a Swift 4 String now has a direct count property that gives the character count:
// Swift 4
let spString = "Stack"
spString.count // 5
Examples for String and SubString.
String
Swift 4 String now directly get Element that gives the first character of String: (string.characters.first)
let spString = "Stack"
let firstElement = spString.first //S
SubString
Using SubString get first character.
let spstring = "Welcome"
let indexStartOfText = spstring.index(spstring.startIndex, offsetBy: 1)
let sub = spstring.substring(to: indexStartOfText)
print(sub) //W
That warning is just a top of the iceberg, there were a loot of string changes, strings are again a collection of characters, but we got soemthing new and cool, subStrings :)
This is a great read about this:
https://useyourloaf.com/blog/updating-strings-for-swift-4/
Just remove characters
For example:
stringValue.characters.count
to
stringValue.count
You can also use this code for dictionary grouping without using { $0.characters.first! }.
let cities = ["Shanghai": 24_256_800, "Karachi": 23_500_000, "Beijing": 21_516_000, "Seoul": 9_995_000]
let groupedCities = Dictionary(grouping: cities.keys) { $0.first! }
print(groupedCities)
func validatePhoneNumber(number:String) -> Bool{
if number.count < 10. //deprecated ->(number.characters.count)
{
return false;
}else{
return true;
}
}
You use directly .count and characters is deprecated.
I have asked a question before about this program, but it seems that not all problems are resolved. I am currently experiencing an error that states: "Cannot convert value of type 'String' to expected argument type '_Element' (aka 'Character') on the "guard let indexInWord" line:
guard let letterIndex = letters.indexOf(sender)
else { return }
let letter = letterArray[letterIndex]
guard let indexInWord = word.characters.indexOf(letter)
else {
print("no such letter in this word")
return
}
// since we have spaces between dashes, we need to calc index this way
let indexInDashedString = indexInWord * 2
var dashString = wordLabel.text
dashString[indexInDashedString] = letter
wordLabel.text = dashString
I tried converting the String 'letter' to Character but it only resulted in more errors. I was wondering how I can possibly convert String to argument type "_Element." Please help.
It is hard to treat a string like a list in swift, mostly because the String.characters is not a typical array. Running a for loop on that works, but if you are looking for a specific character given an index, it is a bit more difficult. What I like doing is adding this function to the string class.
extenstion String {
func getChars() -> [String] {
var chars:[String] = []
for char in characters {
chars.append(String(char))
}
return chars
}
}
I would use this to define a variable when you receive input, then check this instead of String.characters
Hello I have a for in loop where elements is the variable being changed and in this case "elements" is a string but there is a corresponding variable out side of the for in loop that has the same name as the string called elements. So what I mean is out side there is a Var time = [some,text,words] and theres a for in loop that calls a STRING named "time" and I would like to know how to convert the string in the for in loop into the variable by some how taking off the "'s (not that simple I know) without specifically saying "time"(the variable) but instead converting the "elements"(which is the string 'time') string into the variable. I hope I was clear enough if I'm not making sense I'll try again.
You cannot refer to local variables dynamically by their names in Swift. This would break a lot of compiler optimizations as well as type safety if you could.
You can refer to object properties by their names if the class conforms to key-value coding. For example:
class X : NSObject {
let time = ["some", "text", "words"]
func readWordsFromProp(name: String) -> String {
guard let list = self.valueForKey(name) as? [String] else {
return ""
}
var result = ""
for word in list {
result += word
}
return result
}
}
let x = X()
print(x.readWordsFromProp("time"))
In general, there are better ways to do things in Swift using closures that don't rely on fragile name-matching. But KVC can be a very powerful tool
Is there a function to capitalize each word in a string or is this a manual process?
For e.g. "bob is tall"
And I would like "Bob Is Tall"
Surely there is something and none of the Swift IOS answers I have found seemed to cover this.
Are you looking for capitalizedString
Discussion
A string with the first character in each word changed to its corresponding uppercase value, and all remaining characters set to their corresponding lowercase values.
and/or capitalizedStringWithLocale(_:)
Returns a capitalized representation of the receiver using the specified locale.
For strings presented to users, pass the current locale ([NSLocale currentLocale]). To use the system locale, pass nil.
Swift 3:
var lowercased = "hello there"
var stringCapitalized = lowercased.capitalized
//prints: "Hello There"
Since iOS 9 a localised capitalization function is available as capitalised letters may differ in languages.
if #available(iOS 9.0, *) {
"istanbul".localizedCapitalizedString
// In Turkish: "İstanbul"
}
An example of the answer provided above.
var sentenceToCap = "this is a sentence."
println(sentenceToCap.capitalizedStringWithLocale(NSLocale.currentLocale()) )
End result is a string "This Is A Sentence"
For Swift 3 it has been changed to capitalized .
Discussion
This property performs the canonical (non-localized) mapping. It is suitable for programming operations that require stable results not depending on the current locale.
A capitalized string is a string with the first character in each word changed to its corresponding uppercase value, and all remaining characters set to their corresponding lowercase values. A “word” is any sequence of characters delimited by spaces, tabs, or line terminators (listed under getLineStart(_:end:contentsEnd:for:)). Some common word delimiting punctuation isn’t considered, so this property may not generally produce the desired results for multiword strings.
Case transformations aren’t guaranteed to be symmetrical or to produce strings of the same lengths as the originals. See lowercased for an example.
There is a built in function for that
nameOfString.capitalizedString
This will capitalize every word of string. To capitalize only the first letter you can use:
nameOfString.replaceRange(nameOfString.startIndex...nameOfString.startIndex, with: String(nameOfString[nameOfString.startIndex]).capitalizedString)
Older Thread
Here is what I came up with that seems to work but I am open to anything that is better.
func firstCharacterUpperCase(sentenceToCap:String) -> String {
//break it into an array by delimiting the sentence using a space
var breakupSentence = sentenceToCap.componentsSeparatedByString(" ")
var newSentence = ""
//Loop the array and concatinate the capitalized word into a variable.
for wordInSentence in breakupSentence {
newSentence = "\(newSentence) \(wordInSentence.capitalizedString)"
}
// send it back up.
return newSentence
}
or if I want to use this as an extension of the string class.
extension String {
var capitalizeEachWord:String {
//break it into an array by delimiting the sentence using a space
var breakupSentence = self.componentsSeparatedByString(" ")
var newSentence = ""
//Loop the array and concatinate the capitalized word into a variable.
for wordInSentence in breakupSentence {
newSentence = "\(newSentence) \(wordInSentence.capitalizedString)"
}
// send it back up.
return newSentence
}
}
Again, anything better is welcome.
Swift 5 version of Christopher Wade's answer
let str = "my string"
let result = str.capitalized(with: NSLocale.current)
print(result) // prints My String