PHP: 我的代码有错误，主要是 mysqli_fetch_assoc() 需要参数 mysqli_result，给定的对象

Question

有人可以帮我修复我的代码吗？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，但出现了一些错误。

<?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();
?>

我得到的输出：

Warning: mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement in /****/login3.php on line 27
Connection successful
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, object given in /****/login3.php on line 37
Login as admin failed!

Answer 1

你搞混了。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

您必须绑定列，因为它必须匹配需要的字段数：

看看这个：

<?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

谢谢大家！现在可以用了！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 class 没有为其定义方法 fetch_assoc()。

$stmt=$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: 我的代码有错误，主要是 mysqli_fetch_assoc() 需要参数 mysqli_result，给定的对象

PHP: Errors in my code, mainly mysqli_fetch_assoc() expects parameter mysqli_result, object given

php

mysqli