Question

I have a custom UITableViewCell with two labels (UILabel). The table cells are used to display information / text. Inside some of these cells (not all) there is text in this way set:

cell.myTextlabel.text = @"http://www.google.de"

Now I want if I click this text / link, a safari webbrowser should open this webpage. How can I do this?

Best Regards Tim.

Answer 1

Set userInteractionEnabled to YES of your label and add a gesture recognizer to it:

myLabel.userInteractionEnabled = YES;

UITapGestureRecognizer *gestureRec = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(openUrl:)];
gestureRec.numberOfTouchesRequired = 1;
gestureRec.numberOfTapsRequired = 1;
[myLabel addGestureRecognizer:gestureRec];
[gestureRec release];

Then implement the action method:

- (void)openUrl:(id)sender
{
    UIGestureRecognizer *rec = (UIGestureRecognizer *)sender;

    id hitLabel = [self.view hitTest:[rec locationInView:self.view] withEvent:UIEventTypeTouches];

    if ([hitLabel isKindOfClass:[UILabel class]]) {
        [[UIApplication sharedApplication] openURL:[NSURL URLWithString:((UILabel *)hitLabel).text]];
    }
}