使用 PDO Prepared Statement 并递增列值

Using PDO Prepared Statement and Incrementing a column value(使用 PDO Prepared Statement 并递增列值)

问题描述

我已阅读此线程:问题递增MySQL/PHP 中的一个字段，带有准备好的语句，但没有看到我的问题的答案.

I've read this thread: Issues incrementing a field in MySQL/PHP with prepared statements but didn't see the answer to my problem.

PDOStatement Object
(

    [queryString] => UPDATE user_alerts SET notif = ? + 2 WHERE ( user_id = ? )   

)

$stmt->execute( array( 'notif', '1' ) );

据我所知，这一切都是正确的.

As far as I can tell, all this is correct.

当上面的代码执行时，不管notif列的值是什么，它都会将notif列设置为2.就好像 SQL 读起来像 SET notif = 2 而不是 SET notif = notif + 2

When the above code executes, it sets the notif column equal to 2 irregardless of what the value of the notif column is. It's as if the SQL is reading like SET notif = 2 instead of SET notif = notif + 2

我一直无法弄清楚，非常感谢您的帮助！

I haven't been able to figure it out and would really appreciate help!

推荐答案

$sql = 'UPDATE user_alerts SET notif = notif + 2 WHERE ( user_id = :userid )';
$prepStatement = $pdo->prepare( $sql );
$prepStatement->execute(array(':userid' => $userid));

您不能将列名绑定到准备好的语句.

You can't bind a column name to a prepared statement.

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