CODEWITHSUNDEEP
swiftxcode
August Kimo
August Kimo

Reputation: 1771

Format String left of multiple characters in Swift 5?

I have some Strings that vary in length but always end in "listing(number)"

myString = 9AMnep8MAziUCK7VwKF51mXZ2listing28

.

I want to get the String without "listing(number)":

9AMnep8MAziUCK7VwKF51mXZ2

.

Methods I've tried such as .index(of: ) only let you format based off one character. Any simple solutions?

Upvotes: 1

Views: 371

Answers (3)

Abiola Akinnubi
Abiola Akinnubi

Reputation: 313

So to solve your issue find the code below with few comments written to try and explain each steps have taken. kindly note i have modified or arrived at this solution using this links as a guide.

  1. https://stackoverflow.com/a/40070835/6596443

  2. https://www.dotnetperls.com/substring-swift

    extension String {
    //
    // Paramter  inputString: This is the string you want to manipulate
    // Paramter- startStringOfUnwanted: This is the string you want to start the removal or replacement from
    //return : The expected output you want but can be emptystring if unable to

    static func trimUnWantedEndingString(inputString: String,startStringOfUnwanted: String) -> String{
        //Output string
        var outputString: String?
        //Getting the range based on the string content
        if let range = myString.range(of: startStringOfUnwanted) {
            //Get the lowerbound of the range
            let lower = range.lowerBound
            //Get the upperbound of the range
            let upper = range.upperBound
            //Get the integer position of the start index of the unwanted string i added plus one to ensure it starts from the right position
            let startPos = Int(myString.distance(from: myString.startIndex, to: lower))+1
            //Get the integer position of the end index of the unwanted string i added plus one to ensure it starts from the right position
            let endPos =  Int(myString.distance(from: myString.startIndex, to: upper))+1
            //Substract the start int from the end int to get the integer value that will be used to get the last string i want to stop trimming at
            let endOffsetBy = endPos-startPos
            //get thes string char ranges of values
            let result = myString.index(myString.startIndex, offsetBy: 0)..<myString.index(myString.endIndex, offsetBy: -endOffsetBy)
            //converts the results to string or get the string representation of the result and then assign it to the OutputString
            outputString = String(myString[result]);
        }
        return outputString ?? "";
    }
}
let myString = "9AMnep8MAziUCK7VwKF51mXZ2listing28"
String.trimUnWantedEndingString(inputString: myString, startStringOfUnwanted:"listing")

Upvotes: -1

vadian
vadian

Reputation: 285160

A possible solution is to search for the substring with Regular Expression and remove the result (replace it with empty string)

let myString = "9AMnep8MAziUCK7VwKF51mXZ2listing28"
let trimmedString = myString.replacingOccurrences(of: "listing\\d+$", with: "", options: .regularExpression)
  • \\d+ searches for one ore more digits
  • $ represents the end of the string

Alternatively without creating a new string

var myString = "9AMnep8MAziUCK7VwKF51mXZ2listing28"
if let range = myString.range(of: "listing\\d+$", options: .regularExpression) {
    myString.removeSubrange(range)
}

Upvotes: 2

Joakim Danielson
Joakim Danielson

Reputation: 52013

Another option is to split the string in parts with "listing" as separator

let result = myString.components(separatedBy: "listing").first

Upvotes: 1

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