php - Uncaught Error : Call to a member function query() on null

Question

我在下面的代码中卡住了几个小时。我不知道如何修复这个错误。

Notice: Undefined variable: mysqli in D:\xampp\htdocs\recon\register.php on line 19

Fatal error: Uncaught Error: Call to a member function query() on null in D:\xampp\htdocs\recon\register.php:19 Stack trace: #0 {main} thrown in D:\xampp\htdocs\recon\register.php on line 19

<?php

$conn = new mysqli('localhost', 'root', '', 'user');

if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$uname = $_POST['uname'];
$psw = $_POST['psw'];
$options = [
'cost' => 12,];
$hashedpassword= password_hash($psw, PASSWORD_BCRYPT, $options);

$result = $mysqli->query("SELECT username FROM registration WHERE username = '$uname'"); 
$row_count = $result->num_rows; 
if($row_count == 1) { 
    echo 'User already exists, try another one.'; } 
else {

    $query = "INSERT INTO user (username, password) VALUES(?, ?)";
    $statement = $mysqli->prepare($query);

    $statement->bind_param('ss', $uname, $hashedpassword);

    if($statement->execute())
    {
         print 'Success! Last inserted record : ' .$statement->insert_id .'<br />'; 
    }
    else
    {
         die('Error : ('. $mysqli->errno .') '. $mysqli->error);
    }
    $statement->close();

}

?>

Answer 1

您正在声明名为 $conn 的 mysqli 实例。这代表您与数据库的连接。您应该在变量 $conn 上调用方法，而不是在(未定义)变量 $mysqli 上调用方法。所以即。你的第 19 行应该是:

$result = $conn->query("SELECT username FROM registration WHERE username = '$uname'")

此外，为了防止查询/网页上的 SQL 注入(inject)，您应该使用 prepared statements任何地方(包括SELECT)。