How to make a UILabel clickable?

Question

I would like to make a UILabel clickable.

I have tried this, but it doesn't work:

class DetailViewController: UIViewController {

    @IBOutlet weak var tripDetails: UILabel!

    override func viewDidLoad() {
        super.viewDidLoad()
        ...
        let tap = UITapGestureRecognizer(target: self, action: Selector("tapFunction:"))
        tripDetails.addGestureRecognizer(tap)
    }

    func tapFunction(sender:UITapGestureRecognizer) {
        print("tap working")
    }
}

Answer 1

Have you tried to set isUserInteractionEnabled to true on the tripDetails label? This should work.

Answer 2

Swift 3 Update

Replace

Selector("tapFunction:")

with

#selector(DetailViewController.tapFunction)

Example:

class DetailViewController: UIViewController {

    @IBOutlet weak var tripDetails: UILabel!

    override func viewDidLoad() {
        super.viewDidLoad()
        ...

        let tap = UITapGestureRecognizer(target: self, action: #selector(DetailViewController.tapFunction))
        tripDetails.isUserInteractionEnabled = true
        tripDetails.addGestureRecognizer(tap)
    }

    @objc
    func tapFunction(sender:UITapGestureRecognizer) {
        print("tap working")
    }
}

Answer 3

SWIFT 4 Update

 @IBOutlet weak var tripDetails: UILabel!

 override func viewDidLoad() {
    super.viewDidLoad()

    let tap = UITapGestureRecognizer(target: self, action: #selector(GameViewController.tapFunction))
    tripDetails.isUserInteractionEnabled = true
    tripDetails.addGestureRecognizer(tap)
}

@objc func tapFunction(sender:UITapGestureRecognizer) {

    print("tap working")
}

Answer 4

Swift 5

Similar to @liorco, but need to replace @objc with @IBAction.

class DetailViewController: UIViewController {

    @IBOutlet weak var tripDetails: UILabel!

    override func viewDidLoad() {
        super.viewDidLoad()
        ...

        let tap = UITapGestureRecognizer(target: self, action: #selector(DetailViewController.tapFunction))
        tripDetails.isUserInteractionEnabled = true
        tripDetails.addGestureRecognizer(tap)
    }

    @IBAction func tapFunction(sender: UITapGestureRecognizer) {
        print("tap working")
    }
}

This is working on Xcode 10.2.

Answer 5

Swift 3 Update

yourLabel.isUserInteractionEnabled = true

Answer 6

Good and convenient solution:

In your ViewController:

@IBOutlet weak var label: LabelButton!

override func viewDidLoad() {
    super.viewDidLoad()

    self.label.onClick = {
        // TODO
    }
}

You can place this in your ViewController or in another .swift file(e.g. CustomView.swift):

@IBDesignable class LabelButton: UILabel {
    var onClick: () -> Void = {}
    override func touchesEnded(_ touches: Set<UITouch>, with event: UIEvent?) {
        onClick()
    }
}

In Storyboard select Label and on right pane in "Identity Inspector" in field class select LabelButton.

Don't forget to enable in Label Attribute Inspector "User Interaction Enabled"

Answer 7

You need to enable the user interaction of that label.....

For e.g

yourLabel.userInteractionEnabled = true

Answer 8

For swift 3.0 You can also change gesture long press time duration

label.isUserInteractionEnabled = true
let longPress:UILongPressGestureRecognizer = UILongPressGestureRecognizer.init(target: self, action: #selector(userDragged(gesture:))) 
longPress.minimumPressDuration = 0.2
label.addGestureRecognizer(longPress)

Answer 9

Thanks researcher

Here's my solution for programmatic user interface using UIKit.

I've tried it only on Swift 5. And It worked.

Fun fact is you don't have to set isUserInteractionEnabled = true explicitly.

import UIKit

open class LabelButon: UILabel {
    var onClick: () -> Void = {}
    
    public override init(frame: CGRect) {
        super.init(frame: frame)
        isUserInteractionEnabled = true
    }
    
    public required init?(coder: NSCoder) {
        super.init(coder: coder)
    }
    
    public convenience init() {
        self.init(frame: .zero)
    }
    
    open override func touchesEnded(_ touches: Set<UITouch>, with event: UIEvent?) {
        onClick()
    }
}

Uses:

override func viewDidLoad() {
    super.viewDidLoad()
    
    let label = LabelButton()
    label.text = "Label"
    label.onClick = {
        // TODO
    }
}

Don't forget to set constraints. Otherwise it won't appear on view.

Answer 10

Pretty easy to overlook like I did, but don't forget to use UITapGestureRecognizer rather than UIGestureRecognizer.

Answer 11

On top of all of the other answers, this depends on where the label is, it might be behind some subviews. You might think you tap on the label but maybe click the top view. To solve this you can bring the label view to the front with the following line.

self.view.bringSubviewToFront(lblView)

Answer 12

As described in the above solution you should enable the user interaction first and add the tap gesture

this code has been tested using

Swift4 - Xcode 9.2

yourlabel.isUserInteractionEnabled = true
yourlabel.addGestureRecognizer(UITapGestureRecognizer(){
                //TODO 
            })