Question: So i've been trying to get one record and edit it by pressing the button that is created in that table, but honestly, i have no idea how to do that. Dx Can anyone help me with this? (Yes, i want the button created for each record. You know, so at the end of every row in the table, every record will have it's own button.)

```
while($rArt = mysqli_fetch_array($sql)){
 echo '<tr><td>' . $rArt['ArtID'] . '</td>';
 echo '<td>' . $rArt['FiliaalID'] . '</td>';
 echo '<td>' . $rArt['Productnaam'] . '</td>';
 echo '<td>' . $rArt['Inkoopprijs'] . '</td>';
 echo '<td>' . $rArt['Voorraad'] . '</td>';
 echo '<td>' . $rArt['Min_voorraad'] . '</td>';
 $voo = $rArt['Voorraad'];
 $minvoo = $rArt['Min_voorraad'];
 $nodig = $minvoo * 2 - $voo;
 $chosen = $rArt['ArtID'];
 echo '<td>' . $nodig . '</td>';
 echo '<td><input type="submit" name="bestel" value="Bestel"></td></tr>';

 if(isset($_GET['bestel'])){
  $query = mysqli_query($mysql, "
   UPDATE artikel a, voorraad v
   SET v.voorraad = v.voorraad + '$nodig'
   WHERE a.artid = v.artid
   AND v.voorraad = '$chosen'");
 }
}

```

Answer: 
I was struggling with this issue for several hours, i have read every relevant topic and found out that the error was caused because under the privacy settings of my device, the camera access to my app was blocked!!! I have never denied access to camera and i don't know how it was blocked but that was the problem!
Calling this method worked for me. Place it after presenting your view.

```
[yourViewBeingPresented.view layoutIfNeeded];

```