相关疑难解决方法(0)

我试图从MySQL表中选择数据,但我收到以下错误消息之一:

mysql_fetch_array()期望参数1是资源,给定布尔值

要么

mysqli_fetch_array()期望参数1为mysqli_result,给定布尔值

要么

在布尔/非对象上调用成员函数fetch_array()

这是我的代码:

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

$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');

while($row = mysql_fetch_array($result)) {
    echo $row['FirstName'];
}

这同样适用于代码

$result = mysqli_query($mysqli, 'SELECT ...');
// mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
while( $row=mysqli_fetch_array($result) ) {
    ...

和

$result = $mysqli->query($mysqli, 'SELECT ...');
// Call to a member function fetch_assoc() on a non-object
while( $row=$result->fetch_assoc($result) ) {
    ...

和

$result = $pdo->query('SELECT ...', PDO::FETCH_ASSOC);
// Invalid …

可能重复:
PHP错误:mysql_fetch_array()期望参数1是资源,给定布尔值

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

注意:未定义的变量:第30行/Library/WebServer/Documents/ArturoLuna_Final/loginCheck.php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？