使用MYSQLi检查数据库中是否存在用户名/激活了用户名

Question

$username = mysqli_real_escape_string($_POST['username']);

$result = mysqli_query($connect, "SELECT * FROM account WHERE username='".$username."' AND activate='0'");

while($row = mysqli_fetch_assoc($result)) {
    echo 'Account not activated.';
}

我是MYSQLi的新手，如果用户输入错误的用户名并且未激活用户名，我需要如何显示错误的帮助。 到目前为止，我已经获得了上面的代码。

Answer 1

尝试这个

  $username = mysqli_real_escape_string($_POST['username']);

  $result = mysqli_query($connect, "SELECT * FROM account WHERE username='".$username."' AND activate='0'");

  if ($result->num_rows) {
echo "you are activated !!";
                          } 

   else
  {
  // do what u like here if not activated
    }