title: Insert an array and show value in PhpMyAdmin
body: <p>I am using checkboxes to see if a user wants "Breakfast,Dinner and Supper"</p>
  
  <p>This is my code for that </p>
  
  <pre><code>&lt;div id="checkboxesFood"&gt;
                  &lt;p&gt;Please select if you want breakfast, dinner and supper&lt;/p&gt;
                  &lt;input type="checkbox" class="food" name="foodboxes[]" value="Breakfast"/&gt;Breakfast
                  &lt;input type="checkbox" class="food" style="margin:0 0 0 19%;" name="foodboxes[]" value="Dinner"/&gt;Dinner
                  &lt;input type="checkbox" class="food" style="margin:0 0 0 19%;" name="foodboxes[]" value="Supper"/&gt;Supper
              &lt;/div&gt;
  </code></pre>
  
  <p>I'm using PHP to connect to a DB and store the value there.</p>
  
  <p>This is my PHP code</p>
  
  <pre><code>$username = $_POST['userName']; //Username
  $view = $_POST['roomChoice'];   //View Of Room
  $paymentMethod = $_POST['payment']; //How user want's to pay
  $selectedFood = $_POST['foodboxes'];    //Food they have selected(Breakfast, Dinner or Supper)
  //$foodData = array($selectedFood);
  $con = new mysqli('localhost','username','mypw','DB');
  
  if($con-&gt;connect_error){
  die("Connection failed: " .$con-&gt;connect_error);
  }
  //$implodedData = implode(",",$foodData);
  $sql = "INSERT INTO Hotellbokning(Name,Room,Food,Payment) VALUES('$username','$view','$implodedData','$paymentMethod')";
  
  
  if($con-&gt;query($sql) == true){
      echo "You did it John!";
  }
  else{
     echo "Error: " . $sql . "&lt;br/&gt;" .$con-&gt;error;
  }
  </code></pre>
  
  <p>Just ignore the variables that is not an issue. When I insert this array <code>$selectedFood</code> into my DB, I recieve the value "Array" on my Food-field.</p>
  
  <p>Is there a way to see the value of this array(the three values Breakfast,Dinner and Supper) in PhpMyAdmin, or is it only possible to echo them out from my DB?</p>
  
  <p>P.S. I tried using implode but it didn't work. I've also tried <code>$foodData = array($selectedFood)</code> But it didn't work either.</p>