sending information to another view controller - ios

Question

I found a barcode scanner project on git hub that I incorporated into my app link. I am using google books API to get information on the books I scanned.
func getBookInfo(isbn: String) {
guard let url = URL(string: "https://www.googleapis.com/books/v1/volumes?q=isbn13:\(isbn)") else {
print("the url is not valid")
return
}
URLSession.shared.dataTask(with: url, completionHandler: {data, response, error -> Void in
guard error == nil else {
print(response)
print(error!.localizedDescription)
return
}
guard let data = data else {
print("no error but no data")
print(response)
return
}
guard let jsonResult = try? JSONSerialization.jsonObject(with: data, options: []) else {
print("the JSON is not valid")
return
}
if let arrayOfTitles = (jsonResult as AnyObject).value(forKeyPath: "items.volumeInfo.title") as? [String] {
self.BookName.text = "\(arrayOfTitles[0])"
print(self.BookName.text!)
}
if let arrayOfAuthors = (jsonResult as AnyObject).value(forKeyPath: "items.volumeInfo.authors") as? [[String]] {
self.Author.text = "\((arrayOfAuthors[0])[0])"
print(self.Author.text!)
}
if let arrayOfCategories = (jsonResult as AnyObject).value(forKeyPath: "items.volumeInfo.categories") as? [[String]] {
self.Category.text = "\((arrayOfCategories[0])[0])"
print(self.Category.text!)
}
if let arrayOfISBN13 = (jsonResult as AnyObject).value(forKeyPath: "items.volumeInfo.industryIdentifiers.identifier") as? [[String]] {
self.ISBN13.text = "\((arrayOfISBN13[0])[0])"
print(self.ISBN13.text!)
}
if let arrayOfISBN10 = (jsonResult as AnyObject).value(forKeyPath: "items.volumeInfo.industryIdentifiers.identifier") as? [[String]] {
self.ISBN10.text = "\((arrayOfISBN10[0])[1])"
print(self.ISBN10.text!)
}
if let arrayOfFormat = (jsonResult as AnyObject).value(forKeyPath: "items.volumeInfo.printType") as? [String] {
self.CoverType.text = "\(arrayOfFormat[0])"
print(self.CoverType.text!)
}
}).resume()
}
After I scan the books and have received the information, I would like to dismiss the view controller that has the barcode scanner and in the view controller that appears, I would like to display the information of the book I just scanned.
extension MultipleImageViewController: BarcodeScannerCodeDelegate {
func barcodeScanner(_ controller: BarcodeScannerController, didCaptureCode code: String, type: String) {
if code.isEmpty {
let delayTime = DispatchTime.now() + Double(Int64(6 * Double(NSEC_PER_SEC))) / Double(NSEC_PER_SEC)
DispatchQueue.main.asyncAfter(deadline: delayTime) {
controller.resetWithError()
}
}
else{
getBookInfo(isbn: code)
let delayTime = DispatchTime.now() + Double(Int64(6 * Double(NSEC_PER_SEC))) / Double(NSEC_PER_SEC)
DispatchQueue.main.asyncAfter(deadline: delayTime) {
controller.dismiss(animated: true, completion: nil)
}
}
}
}
However, when the barcode scanner view controller is dismissed, I have to exit the app and then come back to the app in order for my information to show in the view controller that I want it to. Without leaving the app and coming back the information that I received from the barcode scanner does not display in the desired view controller.

Answer 1

Your UI is not updating because the method viewDidLoad does not get called when that view controller is returned to since it already loaded. Instead, create a delegate for the parent that will get called when you dismiss the child view controller. Sort of like this:
class ParentViewController: UIViewController, BarcodeDelegate {
func presentChildViewController() {
let childViewController = ChildViewController()
childViewController.delegate = self
present(childViewController, animated: true, completion: nil)
}
func dismissedChildViewController(code: String) {
// update UI
}
}
class ChildViewController: UIViewController {
var delegate: BarcodeDelegate?
func dismiss() {
delegate.dismissedChildViewController(code: getCode())
dismiss(animated: true, completion: nil)
}
}
protocol BarcodeDelegate {
func dismissedChildViewController(code: String)
}
It is difficult to really understand what the problem is, so this may or may not be what you're looking for.