php – 如何在pdo-> query中添加变量值

我想升级我的当前代码,这是经常用PDO注入的sql.

目前我一直坚持在PDO查询中使用变量.

如果我有这样的两个论点

$rowsPerPage = 3;

  // by default we show first page
  $pageNum = 1; 

  if (isset($_GET['page'])) {
     $pageNum = MysqL_real_escape_string($_GET['page']);
  }

  $offset = ($pageNum - 1) * $rowsPerPage;

我有这样的查询

$STH = $DBH->query("SELECT News.ID,LEFT(NewsText,650),Title,AID,Date,imgID," .
        "DATE_FORMAT(Date,'%d.%m.%Y.') as formated_date " .
        "FROM News,Categories,NewsCheck  WHERE Name LIKE '%News - Block%' AND CID=Categories.ID AND JID=News.ID ". 
        "ORDER BY `Date` DESC LIMIT $offset,$rowsPerPage");

PDO在查询ORDER BY的最后一行报告错误
当我用这些线替换
“ORDER BY Date DESC LIMIT3,3”);一切正常.

那么如何在PDO :: query中添加变量值呢？

更新：
感谢回答,我已经像这样更新了我的代码

$STH = $DBH->prepare("SELECT News.ID," .
            "DATE_FORMAT(Date,'%d.%m.%Y.') as formated_date " .
            "FROM News,NewsCheck  WHERE Name LIKE '%News - Block%' AND CID=Categories.ID AND JID=News.ID ". 
            "ORDER BY `Date` DESC LIMIT :offset,:rowsPerPage;");

$STH->bindParam(':offset',$offset,PDO::PARAM_STR);
$STH->bindParam(':rowsPerPage',$rowsPerPage,PDO::PARAM_STR);

$STH->execute();

但是发生了错误：

Fatal error: Uncaught exception ‘PDOException’ with message
‘sqlSTATE[42000]: Syntax error or access violation: 1064 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 ”-3’,‘3” at
line 1’ in /pdo/test.PHP:42 Stack trace: #0
/pdo/test.PHP(42): PDOStatement->execute() #1 {main} thrown in
/pdo/test..

第二次更新
像这样从PARAM_STR更改为PARAM_INT

$STH->bindParam(':offset',PDO::PARAM_INT);
$STH->bindParam(':rowsPerPage',PDO::PARAM_INT);

一切正常.

您希望使用预准备语句和查询参数,如下所示：

$sth = $dbh->prepare('SELECT your_column FROM your_table WHERE column < :parameter');
$sth->bindParam(':parameter',$your_variable,PDO::PARAM_STR);
$sth->execute();

即使您使用的是PDO,直接在查询中使用变量也无法保护您免受sql注入的侵害.参数是防止它们的唯一好方法.

原文链接：https://www.f2er.com/php/138266.html

php – 如何在pdo-> query中添加变量值

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