php - how to retrieve data using two while loops -

the first query works fine , retrieves class id have problem second query not retrieve child name

<select name="child-name" id="schild"> <?php     $username = $_session['username'];     $sql1 = "select class_id class emp_id = '$username'";     $result = mysqli_query($database,$sql) or die(mysqli_error($database));      while($row = mysqli_fetch_array($result))     {         $classid = $row['class_id'];         $sql2 = "select name children class_id = '$classid'";         $result2 = mysqli_query($database,$sql2) or die(mysqli_error($database));         while($row = mysqli_fetch_array($result2)) ?>     <option><?php echo $row['name'];?></option>    <?php          } ?>  </select> 

you using $row in both resultset fetches. therefore destroying outer loop

while testing add these lines, specially if testing on live/hosted server error reporting turned off

<?php     ini_set('display_errors', 1);      ini_set('log_errors',1);      error_reporting(e_all);      mysqli_report(mysqli_report_error | mysqli_report_strict); ?>  <select name="child-name" id="schild"> <?php     $username = $_session['username'];     $sql1 = "select class_id class emp_id = '$username'";     $result = mysqli_query($database,$sql1) or die(mysqli_error($database));     // late breaking fix             ^^^^      while($row = mysqli_fetch_array($result))     {         $classid = $row['class_id'];         $sql2 = "select name children class_id = '$classid'";         $result2 = mysqli_query($database,$sql2) or die(mysqli_error($database));         while($row2 = mysqli_fetch_array($result2)) {     // fix    ^^^^^ ?>     <option value="<?php echo $row2['name'];?>"><?php echo $row2['name'];?></option>   <?php      // fix             ^^^^^         } ?>  </select> <?php     // add terminator first while loop     } ?>