Question

尝试使用我的网络系统中发布的数据更新数据库时遇到问题。我已经创建了一个包含html表单的php文件，就我的订单而言，我可以选择几个产品及其数量。然后我创造了这个：

   <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
    <?php 
        require_once("buzzcafe_fns.php");
     do_html_header('Order form');
    db_connect();
        $otable=$_POST['otable'];
        $title=$_POST['title'];
        $quantity=$_POST['quantity'];
        $datetime=$_SESSION['datetime'];
        $username=$_SESSION['username'];

        if(isset($_SESSION['username']))
                {
                        if( !$otable || !$title || !$quantity )
                        {
                            echo'<font color="red">Please fill in all required fields</font>';
                            exit;
                        }
                }
            $insertOrder = mysql_query('INSERT INTO orders VALUES($oid, $title, $quantity, $sum, $datetime, $username)')or die(mysql_error());

                        echo ('<p>Your order is added</p>');
                        echo ('<p>View your order <a href="vieworder.php?insertOrder='.$insertOrder.'"></a></p>');
                        echo ('<p>Add a new order <a href="addorder-form.php"></a></p>');
                        echo ('<p>Return in main page<a href="members.php"></a></p>');

?>

这段代码有什么问题？

Answer 1

这个声明，实际上是核心，是不正确的

$insertOrder = mysql_query('INSERT INTO orders VALUES($oid, $title, $quantity, $sum, $datetime, $username)')or die(mysql_error());

应该是

$insertOrder = mysql_query("INSERT INTO orders VALUES($oid, $title, $quantity, $sum, $datetime, $username)")or die(mysql_error());

如何从php / html表单中获取数据到数据库？

1 个答案: