警告：mysql_fetch_assoc() 期望参数 1 是给定的资源、对象

发布于 2024-12-21 12:34:31 字数 1033 浏览 0 评论 0原文

我似乎无法弄清楚我做错了什么。因此，当我提交表单时，我收到警告错误并且

注意：未定义变量：/Library/WebServer/Documents/ArturoLuna_Final/loginCheck.php 第 30 行中的 dbusername

$username = $_POST['username'];
$password = $_POST['password']; 

if($username&&$password)
{
    require 'conn.php';
    $query = "SELECT * FROM  users WHERE username='$username'";
    $result = $mysql->query($query) or die(mysqli_error($mysql));
    $numrows = $result->num_rows;
    
    if ($numrows!=0)
    {
        while($row = mysql_fetch_assoc($result))
        { 
            $dbusername = $row['username'];
            $dbpassword = $row['password']; 
        }

        //check to see if they match!
        if($username==$dbusername&&$password==$dbpassword)
        { 
            echo "youre In!";
        }
        else
        echo "incorrect password!";
        
    }
    else
        die("that user is dead");
    
    //echo $numrows;
    
}

else

    echo ("Please Enter Username")

我可能做错了什么？

原文

I cant seem to figure out what I'am doing wrong. So when I submit my form I get Warning error and the

Notice: Undefined variable: dbusername in /Library/WebServer/Documents/ArturoLuna_Final/loginCheck.php on line 30

$username = $_POST['username'];
$password = $_POST['password']; 

if($username&&$password)
{
    require 'conn.php';
    $query = "SELECT * FROM  users WHERE username='$username'";
    $result = $mysql->query($query) or die(mysqli_error($mysql));
    $numrows = $result->num_rows;
    
    if ($numrows!=0)
    {
        while($row = mysql_fetch_assoc($result))
        { 
            $dbusername = $row['username'];
            $dbpassword = $row['password']; 
        }

        //check to see if they match!
        if($username==$dbusername&&$password==$dbpassword)
        { 
            echo "youre In!";
        }
        else
        echo "incorrect password!";
        
    }
    else
        die("that user is dead");
    
    //echo $numrows;
    
}

else

    echo ("Please Enter Username")

what can I be possibly doing wrong?

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瀞厅☆埖开 2024-12-28 12:34:31

更改

while($row = mysql_fetch_assoc($result))

为

while($row = $result->fetch_assoc())

您错过了函数名称中的 i，并且混淆了 OO 和过程式代码，因此您混淆了 mysql_* 结果资源和 mysqli_result对象。

Change

while($row = mysql_fetch_assoc($result))

while($row = $result->fetch_assoc())

You missed an i off the function name, and mixed up OO and procedural style code, so you are mixing up mysql_* result resources and mysqli_result objects.

回复收藏 0 原文

云淡风轻 2024-12-28 12:34:31

您将传统的 PHP MySQL 函数 mysql_* 与 MySQLi 接口混合在一起。在本例中，mysql_fetch_assoc() 需要使用mysql_query() 创建的资源句柄。

要检索正确的结果，您需要使用 MySQLi 的方法：

$row = $result->fetch_assoc();

当您使用它时，您可能需要考虑使代码不易受到 MySQL 注入的影响。正确转义 任何用户提供的输入在将其插入查询字符串之前，或者更好的是，使用 MySQLi 参数绑定工具。

You're mixing traditional PHP MySQL functions mysql_* with the MySQLi interface. In this case, mysql_fetch_assoc() expects a resource handle created with mysql_query().

To retrieve the correct result, you'll need to use MySQLi's method:

$row = $result->fetch_assoc();

And while you're at it, you might want to consider making the code less susceptible to MySQL injection. Properly escape any user provided input before interpolating it into a query string, or better yet, use MySQLi parameter binding facilities.

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