我如何使用ajax从PHP中的一个表中显示的两个表的下拉列表中获取数据

Question

This is code of my Table in PHP. I want to fetch other details from different 2 tables using AJAX in PHP. i am using mysql.I want to show details in table when i select any value from dropdown. Help me please.. I tried so much time but i am not success.

My code is given below.

    <td>
    <?php
    $query = "select * from table_name  order by name ASC ";
    $result= mysql_query($query);
    echo '<select name="id">';
    echo '<option value="">---Select Employee---</option>';
    while ($row=mysql_fetch_array($result))
        {
    ?>
<option value="<?php echo $row['id']; ?>"> <?php echo  ucfirst($row['name']); ?></option>                               
    <?php
         }
         echo "</select>";
    ?>
</td> 

Answer 1

If you want to fetch your selected value from dropdown. then this will help you.

<?php
   $sql= "select * from table_name";
   $res=mysql_query($sql);
   while($row=mysql_fetch_array($res))
   {
?>
     <select name="xyz">
       <option value=""> --- Select ---</option>
<?php
       $sql2="select * from table_name"; 
       $res2=mysql_query($sql2);
       while($row2=mysql_fetch_array($res2))
       {
         if($row2['column_name']==$row['column_name'])
         {
           $selected='selected="selected"';
         } else {
           $selected='';
         }  
?>
          <option value="<?php echo $row2['column_name']; ?>" 
            <?php echo $selected; ?>><?php echo $row2['column_name'] ?>
          </option>
<?php
       }
?>
    </select>
<?php
  }
?>

I think it may help you