您的SQL语法有错误; 检查与MySQL服务器版本对应的手册,以便在第2行的''''''附近使用正确的语法

Question

我在MySQL中遇到错误:

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 '''')' at line 2'.

HTML代码:

<form action="read_message.php" method="post">
  <table class="form_table">
    <tr>
      <td style="font-weight:bold;">Subject:</td>
      <td><input style=" width:300px" name="form_subject"/></td>
      <td></td>
    </tr>
    <tr>
      <td style="font-weight:bold;">Message:</td>
      <td id="myWordCount">&nbsp;(300 words left)</td>
      <td></td>
    </tr>
    <tr>
      <td><input type="hidden" name="sender_id" value="<?php echo $sender_id?>"></td>
      <td><textarea cols="50" rows="4" name="form_message"></textarea></td>
      <td valign="bottom"><input type="submit" name="submit_message" value="send"></td>
    </tr>
  </table>
</form>

要插入mysql表的代码:

<?php
  include_once"connect_to_mysql.php";
  //submit new message
  if($_POST['submit_message']){

    if($_POST['form_subject']==""){
      $submit_subject="(no subject)";
    }else{
      $submit_subject=$_POST['form_subject'];   
    }
    $submit_message=$_POST['form_message'];
    $sender_id = $_POST['sender_id'];
    if($shortMessagesLeft<1){
      $form_error_message='You have left with '.$shortMessagesLeft.' Short Message. Please purchase it from the <a href="membership.php?id='.$id.'">shop</a>.';
    }
    else if($submit_message==""){
      $form_error_message = 'Please fill in the message before sending.';
    }
    else{
      $message_left = $shortMessagesLeft-1;
      $update_short_message = mysql_query("UPDATE message_count SET short_message = '$message_left' WHERE user_id = '$id'");
      $sql = mysql_query("INSERT INTO private_messages (to_id, from_id, time_sent, subject, message) 
        VALUES('$sender_id', '$id', now(),'$submit_subject','$submit_message')") or die (mysql_error());
    }
  }

?>

这个错误意味着什么,我做错了什么？

Answer 1

在$submitsubject或中有一个单引号$submit_message

为什么这是个问题？

单引号char终止MySQL中的字符串,过去的所有内容都被视为sql命令.你真的不想写那样的sql.充其量,您的应用程序将间歇性地中断(正如您所观察到的那样),最糟糕的是,您刚刚引入了一个巨大的安全漏洞.

想象一下,如果有人提交'); DROP TABLE private_messages;了提交消息.

您的SQL命令将是:

INSERT INTO private_messages (to_id, from_id, time_sent, subject, message) 
        VALUES('sender_id', 'id', now(),'subjet','');

DROP TABLE private_messages;

相反,您需要正确清理您的价值观.

在最低限度你必须运行每个值,mysql_real_escape_string()但你应该真正使用预准备语句.

如果您使用mysql_real_escape_string()的代码如下所示:

if($_POST['submit_message']){

if($_POST['form_subject']==""){
    $submit_subject="(no subject)";
}else{
    $submit_subject=mysql_real_escape_string($_POST['form_subject']); 
}
$submit_message=mysql_real_escape_string($_POST['form_message']);
$sender_id = mysql_real_escape_string($_POST['sender_id']);

这是一篇关于预备语句和PDO 的精彩文章.