How to check validity of URL in Swift?

Question

Trying to make an app launch the default browser to a URL, but only if the URL entered is valid, otherwise it displays a message saying the URL is invalid.

How would

Answer 1

Try this:

func isValid(urlString: String) -> Bool
{
    if let urlComponents = URLComponents.init(string: urlString), urlComponents.host != nil, urlComponents.url != nil
    {
        return true
    }
    return false
}

This simply checks for valid URL components and if the host and url components are not nil. Also, you can just add this to an extensions file

Answer 2

My personal preference is to approach this with an extension, because I like to call the method directly on the string object.

extension String {

    private func matches(pattern: String) -> Bool {
        let regex = try! NSRegularExpression(
            pattern: pattern,
            options: [.caseInsensitive])
        return regex.firstMatch(
            in: self,
            options: [],
            range: NSRange(location: 0, length: utf16.count)) != nil
    }

    func isValidURL() -> Bool {
        guard let url = URL(string: self) else { return false }
        if !UIApplication.shared.canOpenURL(url) {
            return false
        }

        let urlPattern = "^(http|https|ftp)\\://([a-zA-Z0-9\\.\\-]+(\\:[a-zA-Z0-9\\.&%\\$\\-]+)*@)*((25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9])\\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[0-9])|localhost|([a-zA-Z0-9\\-]+\\.)*[a-zA-Z0-9\\-]+\\.(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-zA-Z]{2}))(\\:[0-9]+)*(/($|[a-zA-Z0-9\\.\\,\\?\\'\\\\\\+&%\\$#\\=~_\\-]+))*$"
        return self.matches(pattern: urlPattern)
    }
}

This way it is also extensible with another use-cases, such as isValidEmail, isValidName or whatever your application requires.

Answer 3

You can use the NSURL type (whose constructor returns an optional type) combined with an if-let statement to check the validity of a given URL. In other words, make use of the NSURL failable initializer, a key feature of Swift:

let stringWithPossibleURL: String = self.textField.text // Or another source of text

if let validURL: NSURL = NSURL(string: stringWithPossibleURL) {
    // Successfully constructed an NSURL; open it
    UIApplication.sharedApplication().openURL(validURL)
} else {
    // Initialization failed; alert the user
    let controller: UIAlertController = UIAlertController(title: "Invalid URL", message: "Please try again.", preferredStyle: .Alert)
    controller.addAction(UIAlertAction(title: "OK", style: .Default, handler: nil))

    self.presentViewController(controller, animated: true, completion: nil)
}

Answer 4

Version that works with Swift 4.2 and has reliable URL pattern matching ...

func matches(pattern: String) -> Bool
{
    do
    {
        let regex = try NSRegularExpression(pattern: pattern, options: [.caseInsensitive])
        return regex.firstMatch(in: self, options: [], range: NSRange(location: 0, length: utf16.count)) != nil
    }
    catch
    {
        return false
    }
}


func isValidURL() -> Bool
{
    guard let url = URL(string: self) else { return false }
    if !UIApplication.shared.canOpenURL(url) { return false }

    let urlPattern = "(http|ftp|https):\\/\\/([\\w+?\\.\\w+])+([a-zA-Z0-9\\~\\!\\@\\#\\$\\%\\^\\&\\*\\(\\)_\\-\\=\\+\\\\\\/\\?\\.\\:\\;\\'\\,]*)?"
    return self.matches(pattern: urlPattern)
}

Answer 5

Swift 5.1 Solution

extension String {
    func canOpenUrl() -> Bool {
        guard let url = URL(string: self), UIApplication.shared.canOpenURL(url) else { return false }
        let regEx = "((https|http)://)((\\w|-)+)(([.]|[/])((\\w|-)+))+"
        let predicate = NSPredicate(format:"SELF MATCHES %@", argumentArray:[regEx])
        return predicate.evaluate(with: self)
    }
}

Answer 6

If your goal is to verify if your application can open a URL, here is what you can do. Although safari can open the URL, the website might not exist or it might be down.

`// Swift 5
 func verifyUrl (urlString: String?) -> Bool {
    if let urlString = urlString {
        if let url = NSURL(string: urlString) {
            return UIApplication.shared.canOpenURL(url as URL)
        }
    }
    return false
}
`

As a side note, this does not check whether or not a URL is valid or complete. For example, a call that passes "https://" returns true.