php - MySQLi - 试图获取非对象的属性

Question

所以我试图从我的 MySQL 数据库中获取一些数据，但我不断收到这个错误:

Notice: Trying to get property of non-object on line 28

这是我的代码:

  $uname = $_POST['uname'];
  $upass = $_POST['upass'];

  $servername = "<hostname>";
  $username = "<user>";
  $password = "<pass>";
  $dbname = "<dbname>";

  // Create connection
  $conn = new mysqli($servername, $username, $password);

  // Check connection
  if ($conn->connect_error) {
      die("Connection failed: " . $conn->connect_error);
  }

  $sql = "SELECT username, password FROM users WHERE username='".$uname."' AND password='".$upass."'";
  $result = $conn->query($sql);

  if ($result->num_rows == 1) {
    while($row = $result->fetch_assoc()) {
      $_SESSION['uname'] = $row['username'];
    }
  } else {
    echo "Incorrect Details";
  }

  //close connection
  $conn->close();

Answer 1

嘿，我想你需要添加你的数据库
在你的连接中命名我的 friend 。

你需要改变这个

$conn = new mysqli($servername, $username, $password);

对此

$conn = new mysqli($servername, $username, $password, $dbname);

见 this