Open URL from UIButton in CollectionViewCell

Question

I have a UIButton in my UICollectionViewCell and it's getting data from JSON. Now I need to open a URL from each button (each button have a different url that also comes from JSON).

I managed to open the URL with:

let weburl = "http://example.com"
UIApplication.shared.openURL(URL(string: weburl)!)

But now I need to kinda pass an url to each button. Any ideas of how can i achieve this?

You need to use delegates. Check out this post: http://stackoverflow.com/questions/24099230/delegates-in-swift — Eeshwar, Nov 02 '16 at 17:23

Reynaldo Aguilar · Answer 1 · 2016-11-02T17:52:17.897

1

You can have an array of urls:

let urls = [url1, url2, ...]

And then assign the tag property of each button to the index of its corresponding url. Now you can easily manage what you want:

@IBAction func handleTouch(_ sender: UIButton) {
    // assumes that the buttons' tags start at 0, which isn't a good idea.
    // see @rmaddy comment bellow 
    let url = urls[sender.tag]
    // use the version of the open method shown bellow because the other one becomes deprecated in iOS 10
    UIApplication.shared.open(URL(string: url)!, options: [:], completionHandler: nil)
}

EDIT

Other solution would be to just store the url in the cell itself, and in the button handler open the url corresponding to its cell.

edited Nov 02 '16 at 17:52

answered Nov 02 '16 at 17:27

Reynaldo Aguilar

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A few things to consider. This assumes you set each button's tag from 0 to n - 1 where n is the number of buttons. A tag of 0 is the default for any view with no explicit tag. It's best to avoid relying on a tag value of 0. More importantly, relying on a button's tag is risky. It fails under many circumstances such as if the collection view is dynamic (cells can be added, removed, or reordered). – rmaddy Nov 02 '16 at 17:44
@rmaddy Yeah, you are right. I don't like to use this property too much, but some times it helps. Thanks for the tip "A tag of 0 is the default for any view with no explicit tag. It's best to avoid relying on a tag value of 0." I really helps to avoid bugs – Reynaldo Aguilar Nov 02 '16 at 17:49
Another soln:why can't we make a UIButton subclass and make a property url there. and assign that with each url. – Rajan Maheshwari Nov 02 '16 at 17:53

score 0 · Answer 2 · answered Nov 02 '16 at 18:14

FYI openURL is deprecated in iOS 10. I suggest the following if you need to support older versions of ios:

    let url = URL(string: "alexa://")!
    if #available(iOS 10, *) {
        UIApplication.shared.open(url, options: [:], completionHandler: {
            (success) in
            guard success else {
                //Error here
            }
            //Success here
        })
    } else {
        if let success = UIApplication.shared.openURL(url) {
           //Success here
        } else {
          //Error here
        }
    }

Otherwise just use UIApplication.shared.open. Also I would add a URL field to the data model you are passing to your tableViewCell and just look up the URL from the model.

Open URL from UIButton in CollectionViewCell

2 Answers2