相关疑难解决方法(0)

为什么我不能将表名传递给准备好的PDO语句？

$stmt = $dbh->prepare('SELECT * FROM :table WHERE 1');
if ($stmt->execute(array(':table' => 'users'))) {
    var_dump($stmt->fetchAll());
}

是否有另一种安全的方法将表名插入SQL查询？安全我的意思是我不想这样做

$sql = "SELECT * FROM $table WHERE 1"

I know this problem came multiple times, but i cannot find the mistake in my code. In my update-branch every $stmt->bindValue(...) returns TRUE, but i catch the pdo exception

SQLSTATE[HY093]: Invalid parameter number: parameter was not defined

Inserting a new entry works fine.

In my mysql-database the structure of table 'system' is:

id -> int(11), primary key
computer_name -> varchar(255)
cpu_speed -> int(11)
ram_size -> int(11)
mac_address -> varchar(255)
operating_system -> varchar(255)

My error-throwing code:

// Search for mac_address.
// …

所以我在我的控制器中调用了我的 Laravel 模型上的这个查询

模型：

public static function predikat($masa, $ipk_yudisium){

    $searchFilter = DB::select(DB::raw("SELECT 

m_predikat_yudisium.keterangan
FROM m_predikat_yudisium WHERE
m_predikat_yudisium.masa_awal_berlaku <= :masa and
m_predikat_yudisium.masa_akhir_berlaku >= :masa and
m_predikat_yudisium.range_awal_ipk <= :ipk_yudisium  and
m_predikat_yudisium.range_akhir_ipk >= :ipk_yudisium and
m_predikat_yudisium.group_id = 1"),
            array('masa' => $masa,
                'ipk_yudisium' => $ipk_yudisium
            )
        )
        ;
        foreach ($searchFilter as $predikat){
            return $predikat;
        }
        $predikat = (object) array('keterangan' => "" );
        return $predikat;
    }

控制器 ：

$masa = $request->input('masa');
$ipk_yudisium= $request->input('ipk_yudisium');

$predikat = $data->predikat($masa, $ipk_yudisium); //$data is my model called name

但我无法执行该查询，它收到一条错误消息：

SQLSTATE[HY093]: …