如何从mysql php中检索数据并从下拉列表中显示

Hey guys i am a newbie to php.What problem i am facing is i have created a dropdown which is populated from the data from database using this code.This is working fine for me and it is populating dropdown too

    include('connect.php');
    $query="select * from faculty";
    $result=mysql_query($query);
    while($row = mysql_fetch_assoc($result))
    {$dropdown.="
<option value='{$row['Designation']}'>{$row['Designation']}      </option>";}
    echo "<select>".$dropdown."</select>";

Solution i want is,when a user selects a value from dropdown,result should be retrieved from database and should be displayed in table.Please help me guys

You have to basically :
1) Perform a form sending to some server side script(PHP in your case) when there is a change in the dropdown selection (use onchange event for the dropdown) and fetch the values from the db ,
2) Tell the server side script to spit out an html string which contains the table containing the desired information.3)
3) Output the string on your page.

This will do a page refresh.

If you donot want to have a page refresh, resort to using Ajax. P.S. I recommend using some framework such as jQuery in case you need to use Ajax

What you have to do is something like this:

In your Html:

<select onchange="fetchContent()">
    <option id="1_Designation">abcd</option>
    <option id="2_Designation">1234</option>
    <option id="3_Designation">lkjh</option> 
</select>

In your javascript:

fetchContent()
{
    id = $(this).id;
    $.ajax({
      type: "POST",
      url: "/path/content.php?id="+id,
      success: function(response) {
         $("#tableRow").html(response);
      }
   });
}

In content.php you will have to get the value of id and then do the necessary data retrieval and then return the data.

$id = $_POST['id'];
//retrieve the data to $data
echo $data;