PHP prepare语句返回null

时间:2019-02-01 02:58:05

标签: php mysql xampp

我正在编写一个php文件,以打印mysql数据库中的所有用户。我的问题是在prepare($ sql)和$ statement-> execute之后,语句始终为空。这是我的代码:

public function selectAllUser(){
    $response = array();
    $sql = "SELECT * FROM Tata.users";
    $statement = $this->conn->prepare($sql);
    if(!$statement){
        throw new Exception($statement->error);
    }

    //echo json_encode($statement);
    $statement->execute();





    try {
        throw new Exception($statement->error);
    } catch (Exception $e){}

    $result = $statement->get_result();
    //echo json_encode($result);
    while ($row = $result->fetch_assoc()){
        //echo json_encode($row);
        $response[] = $row;
    }
    return $response;
}

这是我得到的结果:echo $ statement;

{"affected_rows":null,"insert_id":null,"num_rows":null,"param_count":null,"field_count":null,"errno":null,"error":null,"error_list":null,"sqlstate":null,"id":null}

这是我使用此功能的代码:

$file = parse_ini_file("../../../Tata.ini");
$dbhost = trim($file["dbhost"]);
$dbuser = trim($file["dbuser"]);
$dbpass = trim($file["dbpassword"]);
$dbname = trim($file["dbname"]);

require ("secure/access.php");


$access = new access($dbhost,$dbuser,$dbpass,$dbname);
$access->connect();

$users = $access->selectAllUser();
$access->disconnect();
echo json_encode($users);

0 个答案:

没有答案