How to make a UILabel clickable?
IosSwiftUilabelUitapgesturerecognizerIos Problem Overview
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")
}
}
Ios Solutions
Solution 1 - Ios
Have you tried to set isUserInteractionEnabled
to true
on the tripDetails
label? This should work.
Solution 2 - Ios
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")
}
}
Solution 3 - Ios
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")
}
Solution 4 - Ios
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.
Solution 5 - Ios
Swift 3 Update
yourLabel.isUserInteractionEnabled = true
Solution 6 - Ios
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"
Solution 7 - Ios
You need to enable the user interaction of that label.....
For e.g
yourLabel.userInteractionEnabled = true
Solution 8 - Ios
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)
Solution 9 - Ios
Pretty easy to overlook like I did, but don't forget to use UITapGestureRecognizer
rather than UIGestureRecognizer
.
Solution 10 - Ios
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.
Solution 11 - Ios
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
})