Question

我的处理程序中有一个表单：

<form action="../submitcomment.php" method="post">
                <input maxlength=100 size=60 type="text" name="IP" value="' . $ip . '" readonly="readonly" hidden="hidden">
                <input maxlength=100 size=60 type="text" name="BlogId" value="' . $blogId . '" readonly="readonly" hidden="hidden">
                <input maxlength=100 size=60 type="text" name="Date" value="' . $date . '" readonly="readonly" hidden="hidden">         
                <input maxlength=100 size=60 type="text" name="Name" placeholder="Enter Your Name">
                <input maxlength=100 size=60 type="text" name="Email" placeholder="Enter Your Email">
                <input maxlength=100 size=60 type="text" name="Comment" placeholder="Enter Your Comment">
                <br>
                <input type="submit" name="Submit" value="Submit Your Comment">
                </form>

行动是submitcomment.php：

$ip = $_POST['IP'];
$BlogId = $_POST['BlogId'];
$Date = $_POST['Date'];
$Name = $_POST['Name'];
$Email = $_POST['Email'];
$Comment = $_POST['Comment'];

$blog = new Blogs();

if (isset($_POST['Submit'])) 
{
    $addComment = $blog->insertComment($ip, $BlogId, $Date, $Name, $Email, $Comment);
    header('Location: http://www.ryan.archi.dev.netsite.co.uk/Blog?success=1');
}else{
    header('Location: http://www.ryan.archi.dev.netsite.co.uk/Blog?fail=1');
}

引用了我班上的一个函数：

function insertComment($ip, $BlogId, $Date, $Name, $Email, $Comment)
    {
        $query = "INSERT INTO BlogComments (Name, Comment, IPAddress, Email, BlogId, Date) VALUES ('$Name', '$Comment', '$ip', '$Email', '$BlogId', '$Date')";
        $oDatabase = new database;
        $connection = $oDatabase->Connect();
        $result = mysql_query ($query, $connection);
        return $result;
    }

尝试插入不会返回或引发任何错误。据我所知，这应该有效，你能发现我做错了吗？

Answer 1

问题在于名为Date的列 - date是一个保留字（我猜所有已知的RDBMS）。

你必须在你的查询中转义这个词：

INSERT INTO BlogComments (Name, Comment, IPAddress, Email, BlogId, `Date`) VALUES ('$Name', '$Comment', '$ip', '$Email', '$BlogId', '$Date')

此外，您的代码为任何人提供了进行SQL注入攻击的机会，因此您至少应该逃避任何用户输入或更好地使用MySQLi或PDO。

你可以通过php函数http://php.net/mysql_real_escape_string进行转义：

$ip = mysql_real_escape_string($_POST['IP']);
$BlogId = mysql_real_escape_string($_POST['BlogId']);
$Date = mysql_real_escape_string($_POST['Date']);
$Name = mysql_real_escape_string($_POST['Name']);
$Email = mysql_real_escape_string($_POST['Email']);
$Comment = mysql_real_escape_string($_POST['Comment']);

Answer 2

你调试了吗？是否已成功建立与数据库的连接？你没有mysql转义，也许插入查询失败。

MySQL查询不插入行

2 个答案: