title: phpMyAdmin DATE variable to the SELECT command - Fatal error: Uncaught ArgumentCountError
body: <p>I'm new here and I learn PHP and SQL connection. I want to list from the SQL database the room numbers what are not reserved. I create 2 tables ( rooms and reservation ) with fill some exapmles.</p>
  
  <h2>Step 1</h2>
  
  <p>If I put direct DATE for the "end reservation"  DATE("2019-03-11") like this work well and listed the room numbers</p>
  
  <pre><code>function reserv3()
  {
      include "config.php";
      $html = "";
      $sql = 'SELECT roomnr , reserv_start FROM rooms LEFT OUTER JOIN reservation ON reservation.room_id = rooms.room_id AND DATE(reservation.reserv_start)  &gt;= DATE("2019-03-11")';
      $result = mysqli_query($conn, $sql);
      while( $sor = mysqli_fetch_array($result) )
          $html .= "&lt;DIV&gt;".$sor['roomnr'].", ".$sor['reserv_start']."&lt;/DIV&gt;";
      return $html;
  }
  </code></pre>
  
  <h2>Step 2</h2>
  
  <p>I want to replace the fixed DATE("2019-03-11") to a variable.</p>
  
  <pre><code>$today = DATE("2019-03-11");
  function reserv3($today)
  {
      include "config.php";
      $html = "";
      $sql = 'SELECT roomnr , reserv_start FROM rooms LEFT OUTER JOIN reservation ON reservation.room_id = rooms.room_id AND DATE(reservation.reserv_start)  &gt;= '.$today;     
      $result = mysqli_query($conn, $sql);
      while( $sor = mysqli_fetch_array($result) )
          $html .= "&lt;DIV&gt;".$sor['roomnr'].", ".$sor['reserv_start']."&lt;/DIV&gt;";
      return $html;
  }
  </code></pre>
  
  <p>I have in Step 2 the Fatal error
  Fatal error: Uncaught ArgumentCountError: Too few arguments to function reserv3(), 0 passed in</p>
  
  <p>Can you help me how can I put variable to this command ?
  Thank you</p>