CODEWITHSUNDEEP
iosswiftuitableviewuibutton
Kegham K.
Kegham K.

Reputation: 1614

Trouble changing UIButton image

I have a tableView in which i post users feeds and i added a heart button for like in the cell view. I created a class for the cell view and declared my @IBOutlet of the button there. Then in the cellForRowAtIndexPath in the tableview i called the button and made the indexpath.row the tag number of the button itself. Then i added a target with an action to be done and created my @IBAction. Now I'm trying to change the image look of the heart button to red but nothing happens. Is there a problem passing the an UIImage to the button via the sender. I have no errors. And the if like = statement is working correctly. Here is my code:

    override func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
        let cell = tableView.dequeueReusableCellWithIdentifier("cell", forIndexPath: indexPath) as! PostsCellTableViewCell

        cell.heartButton.tag = indexPath.row
        cell.heartButton.addTarget(self, action: "liked:", forControlEvents: .TouchUpInside)
return cell
}

@IBAction func liked (sender: UIButton){

        if like == false{
        sender.imageView?.image = UIImage(contentsOfFile: "red-heart.png")
            like = true
        }
        else{
            sender.imageView?.image = UIImage(contentsOfFile: "white-heart-hi.png")
            like = false
        }
       // self.tableView.reloadData()
    }

Upvotes: 0

Views: 70

Answers (2)

InSearchOf
InSearchOf

Reputation: 170

Try this instead.

    if like == false{
        sender.setImage(UIImage(named: "red-heart.png"), forState: .Normal)
        like = true
    } else {
        sender.setImage.setImage(UIImage(named: "white-heart-hi.png"), forState: .Normal)
        like = false
    }

Upvotes: 2

jbcd13
jbcd13

Reputation: 143

This is a really simple error. You declared your target function as "liked:", but your function is declared as "liked". Notice the lack of a colon the second time. This is critical, and should fix the problem. If this does not fix the problem, make sure you have declared

var like = false

If this still doesn't work, comment and I will continue helping. If it does work, don't forget to check my answer.

Upvotes: -1

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