How to add a character at a particular index in st

问题:

I have a string like this in Swift:

var stringts:String = "3022513240"

If I want to change it to string to something like this: "(302)-251-3240", I want to add the partheses at index 0, how do I do it?

In Objective-C, it is done this way:

 NSMutableString *stringts = "3022513240";
 [stringts insertString:@"(" atIndex:0];

How to do it in Swift?

回答1:

If you are declaring it as NSMutableString then it is possible and you can do it this way:

let str: NSMutableString = "3022513240)"
str.insert("(", at: 0)
print(str)

The output is :

(3022513240)

回答2:

Swift 3

Use the native Swift approach:

var welcome = "hello"

welcome.insert("!", at: welcome.endIndex) // prints hello!
welcome.insert("!", at: welcome.startIndex) // prints !hello
welcome.insert("!", at: welcome.index(before: welcome.endIndex)) // prints hell!o
welcome.insert("!", at: welcome.index(after: welcome.startIndex)) // prints h!ello
welcome.insert("!", at: welcome.index(welcome.startIndex, offsetBy: 3)) // prints hel!lo

If you are interested in learning more about Strings and performance, take a look at @Thomas Deniau's answer down below.

回答3:

var phone= "[phone]"

var indx = phone.index(phone.startIndex,offsetBy: 4)

phone.insert("-", at: indx)

index = phone.index(phone.startIndex, offsetBy: 7)

phone.insert("-", at: indx)

//[phone]

回答4:

var myString = "hell"
let index = 4
let character = "o" as Character

myString.insert(
    character, at:
    myString.index(myString.startIndex, offsetBy: index)
)

print(myString) // "hello"

Careful: make sure that index is bigger than or equal to the size of the string, otherwise you'll get a crash.

回答5:

You can't, because in Swift string indices (String.Index) is defined in terms of Unicode grapheme clusters, so that it handles all the Unicode stuff nicely. So you cannot construct a String.Index from an index directly. You can use advance(theString.startIndex, 3) to look at the clusters making up the string and compute the index corresponding to the third cluster, but caution, this is an O(N) operation.

In your case, it's probably easier to use a string replacement operation.

Check out this blog post for more details.

回答6:

To Display 10 digit phone number into USA Number format (###) ###-#### SWIFT 3

func arrangeUSFormat(strPhone : String)-> String {
    var strUpdated = strPhone
    if strPhone.characters.count == 10 {
        strUpdated.insert("(", at: strUpdated.startIndex)
        strUpdated.insert(")", at: strUpdated.index(strUpdated.startIndex, offsetBy: 4))
        strUpdated.insert(" ", at: strUpdated.index(strUpdated.startIndex, offsetBy: 5))
        strUpdated.insert("-", at: strUpdated.index(strUpdated.startIndex, offsetBy: 9))
    }
    return strUpdated
}

回答7:

You can't use in below Swift 2.0 because String stopped being a collection in Swift 2.0. but in Swift 3 / 4 is no longer necessary now that String is a Collection again. Use native approach of String,Collection.

var stringts:String = "3022513240"
let indexItem = stringts.index(stringts.endIndex, offsetBy: 0)
stringts.insert("0", at: indexItem)
print(stringts) // 30225132400

回答8:

Maybe this extension for Swift 4 will help:

extension String {
    mutating func insert(string:String,ind:Int) {
        self.insert(contentsOf: string, at:string.index(string.startIndex, offsetBy: ind) )
    }
}