如何动态地从公司表中获取相关类别作为选项值？

Here i am selecting company name as an option

Here i need to show relevant categories under company name;

How can i do this?

You can do that by using Jquery and Ajax. This is a simple code snippet to achieve what you want.

First, inside your index.php:

<script 
    type="text/javascript" 
    src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js">
</script>
<script type="text/javascript">

$(document).ready(function(){
    $(".company").change(function(){
        var company = $(this).val();
        var dataString = 'company=' + company;

        $.ajax({
            type: "POST",
            url: "business.php", 
            // You can edit choice2.php for other filename you want
            data: dataString,
            cache: false,
            success: function(html){
                $(".business").html(html);
            } 
        });
    });
});
</script>

Company :
<select name="company" class="company">
    <option selected="selected">--Select Company--</option>
    <?php
        include('db.php');

        $sql= "SELECT Company FROM table_test"; 
        // You should edit table_test for your table name

        $result = mysqli_query($con ,$sql);

        while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
            $company = $row['Company'];
            echo '<option value="'.$company.'">'.$company.'</option>';
        } 
    ?>
</select> 
<br/><br/>

Business :
<select name="business" class="business">
    <option selected="selected">--Select Business--</option>
</select>

Then inside your second file. I called it business.php:

<?php
include('db.php');
if($_POST['company']){
    $company = $_POST['company'];
    $sql= "SELECT Business FROM table_test WHERE Company='$company'";
    $result = mysqli_query($con ,$sql);

    while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
        $business = $row['Business'];
        echo '<option value="'.$business.'">'.$business.'</option>';
    }
}
?>

Then to connect to the database you can use this code snippet. I called this file db.php.

<?php 
$con = mysqli_connect("localhost","root","","database_test"); 
// You should edit database_test with your database name
if (mysqli_connect_errno()){
    echo "Failed to connect to MySQL: ".mysqli_connect_error();
}
?>