Swift 4 'substring(from:)' is deprecated: Please use String slicing subscript with a 'partial range from' operator

Question

i\'ve just converted my little app but i\'ve found this error: \'substring(from:)\' is deprecated: Please use String slicing subscript with a \'partial range from\' operator

Answer 1

If you wish to get substring with specific offset without upper bound do the following:

let index = thisRecord.pubDate.index(thisRecord.pubDate.startIndex, offsetBy: 5)
cell.detailTextLabel?.text = String(thisRecord.pubDate[index...]

This way you create a new String object from your existing String thisRecord.pubDate taking anything from specified index to the end index of original String.

Answer 2

It means you should use the new partial range operator as your upperBound:

let str =  "Hello World !!!"
if let index = str.range(of: "Hello ")?.upperBound {
   let string = String(str[index...])  // "World !!!"
}

In your case

cell.detailTextLabel?.text = String(thisRecord.pubDate[index...]))

Answer 3

In place of substring use suffix. Use like below :

cell.detailTextLabel?.text = String(thisRecord.pubDate.suffix(from: index))

Answer 4

str[..<index]
str[index...]

The code above is "partial range from" Look at this How can I use String slicing subscripts in Swift 4?

Answer 5

Follow the below example to fix this warning: Supporting examples for Swift 3, 4 and 5.

let testStr = “Test Teja”

let finalStr = testStr.substring(to: index) // Swift 3
let finalStr = String(testStr[..<index]) // Swift 4

let finalStr = testStr.substring(from: index) // Swift 3
let finalStr = String(testStr[index...]) // Swift 4 

//Swift 3
let finalStr = testStr.substring(from: index(startIndex, offsetBy: 3)) 

//Swift 4 and 5
let reqIndex = testStr.index(testStr.startIndex, offsetBy: 3)
let finalStr = String(testStr[..<reqIndex])

//**Swift 5.1.3 - usage of index**

let myStr = "Test Teja == iOS"

let startBound1 = String.Index(utf16Offset: 13, in: myStr)
let finalStr1 = String(myStr[startBound1...])// "iOS"

let startBound2 = String.Index(utf16Offset: 5, in: myStr)
let finalStr2 = String(myStr[startBound2..<myStr.endIndex]) //"Teja == iOS"

Answer 6

Most of my strings have A-Za-z and 0-9 content. No need for difficult Index handling. This extension of String is based on the familiar LEFT / MID and RIGHT functions.

extension String {

    // LEFT
    // Returns the specified number of chars from the left of the string
    // let str = "Hello"
    // print(str.left(3))         // Hel
    func left(_ to: Int) -> String {
        return "\(self[..<self.index(startIndex, offsetBy: to)])"
    }

    // RIGHT
    // Returns the specified number of chars from the right of the string
    // let str = "Hello"
    // print(str.left(3))         // llo
    func right(_ from: Int) -> String {
        return "\(self[self.index(startIndex, offsetBy: self.length-from)...])"
    }

    // MID
    // Returns the specified number of chars from the startpoint of the string
    // let str = "Hello"
    // print(str.left(2,amount: 2))         // ll
    func mid(_ from: Int, amount: Int) -> String {
        let x = "\(self[self.index(startIndex, offsetBy: from)...])"
        return x.left(amount)
    }
}