Question

我正在尝试从登录的任何用户获取不同的统计信息。然后回显这些统计信息。登录时我正在使用用户名设置会话。然后我试图从这个用户名中获取id并再次检查具有相同ID的表，然后从该表中获取行。

我的猜测是，因为我在登录时只使用用户名启动会话，所以获取数据的代码不会起作用，因为会话不提供id行。我不确定如何做到这一点，或者我是否正确。

我感谢所有的帮助，我真的被困在这里。

这是我的登录代码：

$query = "SELECT password FROM users WHERE username = '$username'";
    $result = mysqli_query($conn, $query);
    $row = mysqli_fetch_assoc($result);

    //USERDATA
    $dbPassword = $row['password'];

    if (password_verify($password, $dbPassword))
    {
        // echo "The details are correct.";
        $_SESSION['loggedin'] = $username;
    require_once('../../frontend/templates/account-actions.php');
    }
    else
    {
        echo "Passwords do not match!";
    }

这是我从登录用户名的id获取统计数据的代码：

$id = $_SESSION['loggedin'];
$query = "SELECT * FROM stats WHERE id='$id'";
$stmt = mysqli_query($conn, $query);
$result = mysqli_fetch_all($stmt,MYSQLI_ASSOC);

Answer 1

我已将您的代码转换为mysqli准备语句，采用程序方法。

$username = "username_to_search";
$password = "password"; //Password is in plain text since password hash has been used.

$stmt = mysqli_prepare($conn, "SELECT * FROM users WHERE username = ?");

/* bind parameters for markers */
mysqli_stmt_bind_param($stmt, "s", $username); //"s" defines the type of data in the following variables, i.e. String for $username.

/* execute query */
mysqli_stmt_execute($stmt);

mysqli_stmt_store_result($stmt);

$total_rows = mysqli_stmt_num_rows($stmt);

mysqli_stmt_bind_result($stmt, $id_fetched, $username_fetched, $password_fetched); //store every field fetched from the table in sequence. Note that I have added _fetched to make it easier to identify later.

if ($total_rows > 0) {
    while(mysqli_stmt_fetch($stmt)) {
        if (password_verify($password, $password_fetched)) {
            $_SESSION['user_id'] = $id_fetched;
            $_SESSION['username'] = $username_fetched;

            require_once('../../frontend/templates/account-actions.php');
        }
        else {
            echo "Invalid Password!";
        }
    }
} else {
    echo "Invalid Username!";
}

一旦正确存储了SESSION变量，现在您可以轻松找到与此相关的所有内容。用户$_SESSION["user_id"]进行搜索。

如何获取用户名和匹配的ID为Session？

1 个答案: