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How can I create a "hyperlink" using Swift?

I am trying to make individual pieces of text UILabel

clickable. What I'm looking for is commonly referred to as a hyperlink in web development.

<a href="//example.com">Link 1</a>
<a href="//example.com/example">Link 2</a>
<a href="//example.com/other_example">Link 3</a>

Each tag a

is its own UILabel

, and ideally it will open Safari to the specified href

tag when you click on the text between tags.

I've found many resources on how to do this in Objective-C, but they all seem overly complicated and don't translate well to Swift (they fit the Objective-C organizational structure which doesn't work) t works well in Swift and goes against the recommended way of using the language).

Here are a few:

If I had 3 UILabel

s,

Paragraph 1

Point 2

Point 3

then what would be the best "Swift-y" way to make each element open to a different url in Safari?

I could create separate buttons for each, but UILabel

programmatically populated, so I thought that making the text respond to taps might be the best option.

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2 answers


One approach would be something like this:
Assumptions:

  • self.urls is a string array containing the URLs associated with each UILabel

    .
  • Each is UILabel

    tag

    assigned a corresponding index in the array
  • labelTapped:

    installed as a handler touchUpInside

    for labels.


import Foundation
import UIKit

class urltest {

    var urls:[String]

    init() {
        self.urls=[String]()  // Load URLs into here
    }

    @IBAction func labelTapped(sender:UILabel!) {

        let urlIndex=sender.tag;
        if (urlIndex >= 0 && urlIndex < self.urls.count) {
           self.openUrl(self.urls[urlIndex]);
        }

    }

    func openUrl(url:String!) {

        let targetURL=NSURL.URLWithString(url)

        let application=UIApplication.sharedApplication()

        application.openURL(targetURL);

    }
}
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Swift 3 I created a LinkUILabel class in github: https://github.com/jorgecsan/LinkUILabel With this, you only need to add the url, which can be checked as the image is displayed: Example of a label with urlor assign the url variable programmatically:

linkUILabel.url = "www.example.com"

If you want to realize yourself on your own, I found this solution! :)



through:

// This is the label
@IBOutlet weak var label: UILabel!

override func loadView() {
    super.loadView()

    // This is the key
    let tap = UITapGestureRecognizer(target: self, action: #selector(self.onClicLabel(sender:)))
    label.isUserInteractionEnabled = true
    label.addGestureRecognizer(tap)
}

// And that the function :)
func onClicLabel(sender:UITapGestureRecognizer) {
    openUrl("http://www.google.com")
}


func openUrl(urlString:String!) {
    let url = URL(string: urlString)!
    if #available(iOS 10.0, *) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    } else {
        UIApplication.shared.openURL(url)
    }
}

Hope it helps! :)

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