Php 根据id向数据库中插入值

我想根据从下拉列表中选择的值将数据插入表中。 实际上我有两个表，

类别

和

产品

。两者都与

类别ID

链接
现在我有了一个表单，通过这个表单我试图添加一个产品。我有一个类别的下拉列表。我将从下拉列表中选择一个类别，然后将相关产品添加到该类别中

获取表单数据后，我获取了

CategoryID

的表，如下所示：

 include('includes/conn.php');
 if (isset($_POST['submit']))
 { 
     $CategoryName = $_POST['cat'];
     echo $CategoryName; 
     $ProductName = mysql_real_escape_string(htmlspecialchars($_POST['ProductName']));
     $ProductCode = $_POST['ProductCode'];
     $Specification = $_POST['Specification'];
     $Description = $_POST['Description'];
     $CostPrice = $_POST['CostPrice'];
     $DisplayPrice = $_POST['DisplayPrice'];
     $ProductID = $_POST['ProductID'];
     $Productimage = $_POST['ProductImage'];
     $sql = "select * Categories";
     $result = mysql_query ($sql);
     $row = mysql_fetch_array($result);
     $Category_ID = $row['CategoryID'];

这之后我就不知道该怎么办了。除此之外，我的代码成功地插入了记录。 不选择类别的完整代码如下

<?php 
}

include('includes/conn.php');
    if (isset($_POST['submit']))
    { 
        $ProductName = mysql_real_escape_string(htmlspecialchars($_POST['ProductName']));
        $ProductCode = $_POST['ProductCode'];
        $Specification = $_POST['Specification'];
        $Description = $_POST['Description'];
        $CostPrice = $_POST['CostPrice'];
        $DisplayPrice = $_POST['DisplayPrice'];
        $ProductID = $_POST['ProductID'];
        $Productimage = $_POST['ProductImage'];
        if ($ProductName == '' || $ProductCode == ''|| $Specification == '' || $Description == '' || $CostPrice == '' || $DisplayPrice =='')
        {
            echo  "Please fill in all required fields";
            renderForm($ProductID, $ProductName, $ProductCode, $Description, $Specification, $CostPrice, $DisplayPrice,  $error);
        }
        else
        {
            $sql = "INSERT into Products SET ProductName='$ProductName', ProductCode='$ProductCode', Specification ='$Specification', Description = '$Description', CostPrice = $CostPrice, DisplayPrice = $DisplayPrice, ProductImage = '$ProductImage'";
            mysql_query($sql) or die(mysql_error());
            echo " Successfully Added "; 
            //header("Location: view.php"); 
        }
    }
    else
    {
        renderForm('','','','','','','','','');
    }
   ?> 

这就是如何使用插入查询

确保结合使用sprintf和mysql\u real\u escape\u string（）来屏蔽数据库是的，我已经更正了插入查询，但这不是我的问题。我想在cateoryid存在的地方插入数据
$sql = "INSERT INTO Products (ProductName, ProductCode, ...) VALUES ('".$ProductName."', '". $ProductCode ."', ...";