我得到的mysqli的错误和警告(I am getting errors and warnings

更新：

我有一对夫妇，我需要用mysqli的帮助错误和警告：

Fatal error: Call to undefined method mysqli_stmt::get_result() in ... on line 63

在我的代码下面有没有人知道如何将这些警告和错误可以处理？

$query = "SELECT * FROM Teacher WHERE TeacherAlias = ?";
// prepare query
$stmt=$mysqli->prepare($query);
// You only need to call bind_param once
$stmt->bind_param("s",$getid);
// execute query
$stmt->execute();  
//get results
$result = $stmt->get_result(); 

$numrows = mysqli_num_rows($result);
if ($numrows == 0){   

       // don't use $mysqli->prepare here
$query = "SELECT * FROM Teacher WHERE TeacherUsername = ?";
// prepare query
$stmt=$mysqli->prepare($query);
// You only need to call bind_param once
$stmt->bind_param("s",$getuser);
// execute query
$stmt->execute(); 

}

Answer 1:

您需要检索mysqli_result第一对象，喜欢的东西...

$res = $stmt->get_result();

...然后从该对象（不取的行数$stmt ）：

$numrows = mysqli_num_rows($res);

更新： get_result方法是在PHP可用5.3+只，为老版本应该用下面的办法：

// $stmt preparing code goes here...

$stmt->execute();
$stmt->store_result();
$num_rows = $stmt->num_rows;
doSomethingWith($num_rows);

// processing cycle:
$stmt->bind_result($some_param, $another_param);
while ($stmt->fetch()) {
   doSomethingElseWith($some_param, $another_param);
}
$stmt->free_result();
$stmt->close();

作为一个旁注，两项建议：1）它会是可能更快，在这里使用一个单一的查询，并寻找在同时TeacherAlias和TeacherUsername字段值兼而有之（OR运算符，如TeacherAlias = ? OR TeacherUsername = ? ）; 2）它会更容易有明确规定的列（工作SELECT id, TeacherAlias AS alias, TeacherUsername AS username... ），而不是仅仅（ SELECT *在你的查询）。