php - 使用表单php> form> php传递变量时出错

请问这段代码有什么问题？

我用它来向数据库添加一些数据，但是在尝试插入时却得到空的$toid和$toname。

这是表格。变量$toid和$toname在这里可以。

//write new message
if (isset($_GET['action']) && $_GET['action'] == compose) {
        if (isset($_GET['toid'])) {
            $toid = $_GET['toid'];
            $tosql = "select * from authors where id =".$toid."";
            $toquery = mysql_query($tosql,$connection) or die(mysql_error());
            $torow = mysql_fetch_array($toquery);
            $toname = $torow['displayname'];
        if (isset($_GET['subject'])) { $subject = $_GET['subject']; }
        if (isset($_GET['message'])) {
            $message = $_GET['message'];
            echo "<h3>Replying</h3>
            <table>
                <tr>
                    <td colspan='2'>Replying to ".$toname.".</td>
                </tr>
                <tr>
                    <td colspan='2'>".$subject."".nl2br($message)."<br />
                    </td>
                </tr>
            </table><br />Type your answer:<br /><br />";
        } else { echo "New message"; }
        echo "<form action=\"mail.php?action=send\" method=post>
            <table>
                <tr>
                    <td>To:</td><td><input type=\"text\" name=\"to\" size=\"50\" value=\"".$toname."\"></td>
                </tr>
                <tr>
                    <td>Title:</td><td><input type=text name=subject size=50 value=".$subject."></td>
                </tr>
                <tr>
                    <td valign=\"top\">Message:</td><td><textarea  rows=\"10\" cols=\"70\" name=\"message\"></textarea></td>
                </tr>
                <tr>
                    <td align=\"right\" colspan=\"2\"><input id=\"submitstyle\" type=\"submit\" value=\"Enviar Mensagem\"></td>
                </tr>
            </table>
        </form>";
        }
}

这是将消息插入到数据库的代码，$toid和$toname在此处为空。应该从上面的表格中获取它，对吧？

//send message
if (isset($_GET['action']) && $_GET['action'] == send) {

    if ($subject == "" || $message == "") {
        header('Location: mail.php?action=compose&toid='.$toid.'&subject=\''.$subject.'\'&sendpm=false');
        exit();
    }

    $date = DATE(YmdHis);

    echo $userid."from<br />to".$toid."<br />toname".$toname;

    $sendsql = "INSERT INTO mail (sender, reciever, subject, message, created_at, status, sender_deleted, reciever_deleted)
                        VALUES (".$userid.", ".$toid.", ".$subject.", ".$message.", ".$date.",unread, 0, 0)";
    $sendquery = mysql_query($sendsql,$connection) or die(mysql_error());

    echo "<div class=\"alert\" style=\"text-align: center; margin-top: 13px;\"><b>Mensagem particular enviada com sucesso!</b></div>
            <br /><table align=\"center\" width=\"75%\" class=\"sortable\">
                <tr>
                    <td colspan='2' style=\"text-align:center;font-weight:normal;\">Mensagem particular enviada para ".$toname.".</td>
                </tr>
                <tr>
                    <td colspan='2'>
                        Title: ".$subject."
                        Message: ".nl2br($message)."

                    </td>
                </tr>
            </table>";
}

我也因为第2行得到这个sql错误You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' RE: assunto3, 3, 20121017023723,unread, 0, 0)'，因为我提到了空变量。

最佳答案

首先，您需要将这些变量作为隐藏的表单值传递：​​http://www.echoecho.com/htmlforms07.htm

然后，您需要通过$ _POST从表单中获取变量。

尝试$ toid = $ _POST ['toid']和$ toname = $ _POST ['toname']。但是请注意SQL注入：http://php.net/manual/en/security.database.sql-injection.php

不要只是盲目接受$ _POST的值。请务必先进行验证和过滤。

或者，如果用户无法更改toid / toname，为什么不重新查询它们呢？