Question

因此，我从TutorialRepublic复制了一份注册表单和登录表单，当该表单不起作用时，我添加了一些警报以找出未在运行哪些代码。

我最终发现mysqli_prepare部分已被完全跳过：

$sql = "SELECT id, username, password FROM acccounts WHERE username = ?";

    if($stmt = mysqli_prepare($link, $sql)){
        // Bind variables to the prepared statement as parameters
        mysqli_stmt_bind_param($stmt, "s", $param_username);

        echo "<script type='text/javascript'>alert('yes');</script>";

        // Set parameters
        $param_username = $username;

        // Attempt to execute the prepared statement
        if(mysqli_stmt_execute($stmt)){
            // Store result
            mysqli_stmt_store_result($stmt);

            // Check if username exists, if yes then verify password
            if(mysqli_stmt_num_rows($stmt) == 1){                    
                // Bind result variables
                mysqli_stmt_bind_result($stmt, $id, $username, $hashed_password);
                if(mysqli_stmt_fetch($stmt)){
                    if(password_verify($password, $hashed_password)){
                        // Password is correct, so start a new session
                        session_start();

                        // Store data in session variables
                        $_SESSION["loggedin"] = true;
                        $_SESSION["id"] = $id;
                        $_SESSION["username"] = $username;                            

                        // Redirect user to welcome page
                        header("location: index.php");
                    } else{
                        // Display an error message if password is not valid
                        $password_err = "The password you entered was not valid.";
                    }
                }
            } else{
                // Display an error message if username doesn't exist
                $username_err = "No account found with that username.";
            }
            // Close statement
            mysqli_stmt_close($stmt);
        } else{
            echo "Oops! Something went wrong. Please try again later.";
        }
    }
    else{
        echo "<script type='text/javascript'>alert('error');</script>";
    }

我知道最好只是带别人的教程代码，然后逐字使用它，但是我真的只是需要这样做。我对PHP和MySQL还是很陌生，所以我什至不知道从哪里开始尝试调试。

Answer 1

好的，所以本质上我只需要调试更好。在两个用户的评论帮助下，我能够找到错误所在。

我已经做出else的声明，警告说mysqli_prepare已被跳过。在该else语句中，我应该添加echo mysqli_error ($link);，因为这将输出错误信息。

mysqli_prepare被跳过

1 个答案: