How to make a UILabel clickable?

≡放荡痞女 提交于 2019-11-26 06:28:41

问题


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\")
    }
}

回答1:


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




回答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")
    }
}



回答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")
}



回答4:


Swift 3 Update

yourLabel.isUserInteractionEnabled = true



回答5:


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.




回答6:


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

For e.g

yourLabel.userInteractionEnabled = true




回答7:


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"




回答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)



回答9:


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




回答10:


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 
            })


来源:https://stackoverflow.com/questions/33658521/how-to-make-a-uilabel-clickable

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