計算日数は、チェックアウトにチェックインをもとにとどまりますゲストはそれを計算する方法？チェックアウト時間が12時正午である場合

Question

<?php 
    $today = date('Y-m-d'); 
    $sql = "SELECT * FROM roombooking where enddate>='$today'" ; 

    $result1 = $db->query($sql); 
    if ($result1->num_rows > 0){ 
    echo" 
    <div class=table-responsive border > 
     <table class=table> 
     <tr> 

     <th>Room Type</th> 
     <th>Room Number</th> 
     <th> Status </th> 
     <th> Check-in date </th> 
     <th> Check-Out date </th> 
     <th>Stay Time</th> 
     <th colspan='2'>Action </th> 
     </tr>"; 
     // output data of each row 

     // output data of each row 
     while ($row = $result1->fetch_assoc()) 
     { echo "<tr> 
       <td>".$row["roomtype"]."</td> 
       <td>".$row["roomnumber"]."</td> 
       <td>".$row["bkg_status"]."</td> 
       <td>".$d1=$row["bkg_clientintime"]."</td> 
       <td>".$d2=$row["bkg_clientouttime"]."</td> 


       </tr>"; 
      } 
    echo "</table></div>"; 
    }else { echo "<br>"."<div class='alert alert-info col col-md-6 col-md-offset-2'> SORRY NO CHEK OUT FOUND FOR TODAY 

    </div> ";} 
?> 
//here from $d1 which is booking check-in-date and $d2 which is check-out-date 
HELP REQUIRED // important checkout date time is 12.00 noon 
HELP REQUIRED 2// FINDING TOTAL Number of days stay 

any suggestion are greatly appreciated 
tried here date_diff/it again gives hours or days but have to find based on 
12.00 noon important checkout otherwise new day will be added how to achieve this part kindly guide 

Answer 1

で最も簡単な方法は、

SELECT *, 
    DATE_DIFF(DATE(bkg_clientintime), DATE(bkg_clientouttime)) AS no_days 
FROM roombooking