PHP Mysql不接受表名作为变量

Question

mysql无法识别函数中变量中我的表的名称，它会是什么？

我的 PHP 代码:

$TableMaster = "table_name";
function recursiveDelete($id,$db,$table){
    $db_conn = $db;
    $query = $db->query("SELECT * FROM ".$table." WHERE Padre = '".$id."' ");
    if ($query->rowCount()>0) {
        while($current=$query->fetch(PDO::FETCH_ASSOC)) {
            recursiveDelete($current['id'],$db_conn);
        }
    }
    $db->exec("DELETE FROM ".$table." WHERE id = '".$id."' ");
}
recursiveDelete($_POST['id'],$db,$TableMaster);

错误 PHP 日志:

PHP 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 'WHERE Father = '99'' at line 1' in

注意:但是当我在语句中直接写我的mysql表名是没有问题的。

发生了什么事？

Answer 1

在进行递归调用时，您遗漏了 $table 参数。

也不需要$db_conn变量，你可以直接使用$db。

function recursiveDelete($id,$db,$table){
    $query = $db->query("SELECT * FROM ".$table." WHERE Padre = '".$id."' ");
    if ($query->rowCount()>0) {
        while($current=$query->fetch(PDO::FETCH_ASSOC)) {
            recursiveDelete($current['id'],$db,$table);
        }
    }
    $db->exec("DELETE FROM ".$table." WHERE id = '".$id."' ");
}