Question

有人可以帮我修改我的代码吗？我正在尝试访问密码哈希数据库并使用它们来验证用户登录，但是我收到了一些错误。

<?php
$servername="localhost";
$username = "*****";
$password = "*******";
$dbname = "*****";
$conn = new mysqli($servername,$username,$password,$dbname);
if($conn->connect_error){
    die("Connection to database failed: ".$conn->connect_error);
}
$uname=mysqli_real_escape_string($conn, $_POST['entered_username']);
$pw=mysqli_real_escape_string($conn, $_POST['entered_password']);

$stmt=$conn->prepare("SELECT username,password,password_hash FROM users WHERE username=?");
$stmt->bind_param('s',$uname);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($result);
$stmt->fetch();

if(!$stmt){
echo $conn->connect_error();}
if($stmt){
echo 'Connection successful';}

$found=FALSE;

while($row=mysqli_fetch_assoc($stmt))
{
    if($password_verify($pw,$row['password_hash'])) {
        $found=TRUE;
    }
}


if($found){
    echo "You have successfully logged in as ".$uname."!";
}
else {
    echo "Login as ".$uname." failed!";
}
$stmt->close();
$conn->close();
?>

我得到的是输出：

警告：mysqli_stmt :: bind_result（）：绑定变量数没有   匹配预备语句中的字段数   第27行///login3.php
连接成功
警告：mysqli_fetch_assoc（）期望参数1为mysqli_result，   第37行的/****/login3.php中给出的对象
以管理员身份登录失败！

Answer 1

你正在混淆它。试试 -

$stmt=$conn->prepare("SELECT username,password,password_hash FROM users WHERE username=?");
$stmt->bind_param('s',$uname);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($result);


if(!$stmt){
echo $conn->connect_error();}
if($stmt){
echo 'Connection successful';}

$found=FALSE;

while($row=$stmt->fetch_assoc();)
{
    if($password_verify($pw,$row['password_hash'])) {
        $found=TRUE;
    }
}

Answer 2

你必须绑定列，因为它必须匹配需要的字段数：

检查一下：

<?php
$servername="localhost";
$username = "*****";
$password = "*******";
$dbname = "*****";
$conn = new mysqli($servername,$username,$password,$dbname);
if($conn->connect_error){
    die("Connection to database failed: ".$conn->connect_error);
}
$uname=mysqli_real_escape_string($conn, $_POST['entered_username']);
$pw=mysqli_real_escape_string($conn, $_POST['entered_password']);

$stmt=$conn->prepare("SELECT username,password,password_hash FROM users WHERE username=?");
$stmt->bind_param('s',$uname);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($username,$password,$password_hash);
$stmt->fetch();

if(!$stmt){
echo $conn->connect_error();}
if($stmt){
echo 'Connection successful';}

$found=FALSE;

while($row=mysqli_fetch_assoc($stmt))
{
    if($password_verify($pw,$password_hash)) {
        $found=TRUE;
    }
}


if($found){
    echo "You have successfully logged in as ".$uname."!";
}
else {
    echo "Login as ".$uname." failed!";
}
$stmt->close();
$conn->close();
?>

Answer 3

谢谢你们！它现在有效！我更改了bind_result语句并删除了fetch语句。显然，$stmt is of type mysqli_stmt, not mysqli_result和mysqli_stmt类没有为它定义fetch_assoc（）方法。

$stmt=$conn->prepare("SELECT username,password_hash FROM users WHERE username=?");
$stmt->bind_param('s',$uname);
$stmt->execute();
$stmt->store_result();
$stmt->bind_result($user,$password_hash);

$found=FALSE;

while($stmt->fetch())
{
    if(password_verify($pw,$password_hash)) {
        $found=TRUE;
    }
}

PHP：我的代码中的错误，主要是mysqli_fetch_assoc（）需要参数mysqli_result，给出的对象

3 个答案: