快速在 tableview 和 viewController 之间传递值

Question

I know this question is being asked lot of times..trying from last 2 days. Please help me.

I have a custom TableView trying pass few data from Array of this Tableview to next View.

Error : unexpectedly found nil while unwrapping an Optional value.

I tried many ways.. but no success.

EDITED

Finally Solved

override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {


    if segue.identifier == "SegueToQR"
    {
        let controller = segue.destinationViewController as! QRcodeViewController
        if let button = sender as? UIButton {
            let cell = button.superview?.superview?.superview as! PatientTableViewCell
            controller.passName = cell.patientName.text
            controller.passAdress = cell.adress.text
        }
    }

    }

Answer 1

Your logic is wrong. The UITableView only maintains cells that are visible on the screen. So if you try to access a cell that is not visible, it will give nil exception.

First do not use CGPoint to get the index. Assign a tag to each button in your cell, and pass that button in the sender parameter of the performSegue function.

Then you should maintain your patient data in an array. And then access the data using the button.tag as index to the array. You will get your patient model object, from this object you can pass the data to the next view controller.