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Home/ Questions/Q 8976051
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Editorial Team
  • 0
Editorial Team
Asked: June 15, 20262026-06-15T19:01:53+00:00 2026-06-15T19:01:53+00:00

I am trying to populate second dropdown list based on first dropdown list selection

  • 0

I am trying to populate second dropdown list based on first dropdown list selection using Ajax, jQuery, PHP and MySQL. The problem is the options in second dropdown list just appears in one line (all options in one line)!

I was hoping the while loop inside the results.php could handle this but it seems not. How can I fix this issue?

Here is my code:

<html>
<body>
<?php
$mysqli = new mysqli('localhost', 'root', '', 'chain');
if ($mysqli->connect_errno) 
{
die('Unable to connect!');
}
else
{
$query = 'SELECT * FROM Cars';
if ($result = $mysqli->query($query)) 
 {
 if ($result->num_rows > 0) 
    {
 ?> 
 <p>
  Select a Car
<select id="selectCar">
<option value="select">Please Select From The List</option>
    <?php       
     while($row = $result->fetch_assoc()) 
   {
   ?>       
   <option value="<?php echo $row['id']; ?>"><?php echo $row['title']; ?></option>      
  <?php         
}
  ?>
 </select>
 </p>
<select>
    <option value="select">Please Select From The List</option>
    <option id="result"></option>
    </select>
    <?php
    }
    else 
    {
        echo 'No records found!';
    }
    $result->close();
}
else 
{
    echo 'Error in query: $query. '.$mysqli->error;
}
    }
     $mysqli->close();
    ?>
    <script type="text/javascript" src="jquery.js"></script>
    <script type="text/javascript">
        $(document).ready(function()
        {
            $('#selectCar').change(function()
            {
                if($(this).val() == '') return;
                $.get(
                    'results.php',
                    { id : $(this).val() },
                    function(data)
                    {
                        $('#result').html(data);
                    }
                );
            });
        });
    </script>
      </body>
      </html>

In the PHP result I have:

<?php
 $mysqli = new mysqli('localhost', 'root', '', 'chain');
 $resultStr = '';
 $query = 'SELECT * FROM models WHERE carID='.$_GET['id'];
 if ($result = $mysqli->query($query)) 
 {
if ($result->num_rows > 0) 
{
    while($row = $result->fetch_assoc())
    {
  $resultStr.=  '<option    value="'.$row['id'].'">'.$row['model'].'</option>'; 
    }
}
else
{
 $resultStr = 'Nothing found';
}
   }
    echo $resultStr;
   ?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T19:01:55+00:00Added an answer on June 15, 2026 at 7:01 pm

    You should edit your script into this..

    <select id="result">
      <option value="select">Please Select From The List</option>
    </select>
    
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