PHP PDO 语法错误或访问冲突 : 1064

Question

遇到一个我似乎没有看到的问题。

这是我的函数

function addIT($type, $amount, $id) {
$db = new database();
$conn8 = $db->connect();
$addit = $conn8->prepare("UPDATE accountTable SET :type = :type + :amount WHERE ID =     :id");
$addit->execute(array('type'=>$type, 'amount'=>$amount, 'id'=>$id));
}

这是我的电话:

$type = "age";
$amount = 17;
$id = 1;
addIT($type, $amount, $id);

现在执行代码时出现以下错误

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 ''age' = 'age' + '17' WHERE ID = '1'' at line 1' in /Library/WebServer/Documents/me/functions/activity.php:7 Stack trace: #0 /Library/WebServer/Documents/me/functions/activity.php(7): PDOStatement->execute(Array) #1 /Library/WebServer/Documents/me/content/doadd.php(66): addIT('age', 17, '1') #2 /Library/WebServer/Documents/me/main(48): include('/Library/WebSer...') #3 /Library/WebServer/Documents/me/where.php(25): include('/Library/WebSer...') #4 {main} thrown in /Library/WebServer/Documents/me/functions/activity.php on line 7

我确定我的执行或准备语句有问题。我再也看不到了。

Answer 1

查询中不能有可变的列名或表名。使用

UPDATE accountTable 
SET age = age + :amount 
WHERE ID = :id

顺便说一句，您不应该存储一个人的年龄，而应该存储生日。然后你不必更新它。