Question

我正在学习编程，并且正在尝试使用Ruby。我想限制一种功能，该功能允许用户向我发送介绍请求（每30天3个介绍请求）。我不确定是否必须先创建一个方法，例如：

def month
  where("created_at <?", Date.today - 30.days))
end

我不知道该方法是否正确，是否可以将其集成到这段代码中：

def create
  @introduction = CompanyIntroduction.create(intro_params)
  if current_user #Admin can edit the user name and email 
    @introduction.user = current_user
    @introduction.user_name = current_user.full_name
    @introduction.user_email = current_user.email
  end 
  if @introduction.save
    flash[:success] = "Thank you. Your request is being reviewed by TechIreland."
  else
    flash[:error] = @introduction.errors.full_messages
  end
  redirect_back(fallback_location: user_companies_path)
end

Answer 1

您已经关闭。但是，您需要进行时间比较，并且该方法（假设它在模型中）应该是类方法或范围。

@Service
public class UserDetailServiceImpl implements UserDetailsService {

@Autowired
UserDao userDao;

@Override
public UserDetails loadUserByUsername(String username) throws UsernameNotFoundException {
    // TODO Auto-generated method stub
    User user = this.userDao.getByUsername(username);
    if (user == null) {
        System.out.println("User not found! " + username);
        throw new UsernameNotFoundException("User " + username + " was not found in the database");
    }
    System.out.println("Found User: " + user);

    String role = user.getUser_role();
    List<GrantedAuthority> grantList = new ArrayList<GrantedAuthority>();
    if (role != null) {
        GrantedAuthority ga = new SimpleGrantedAuthority(role);
        grantList.add(ga);
    }

    UserDetails userDetails = new org.springframework.security.core.userdetails.User(user.getUsername(),
            user.getEncrypted_password(), grantList);

    return userDetails;
}

}

然后在控制器中，您可以检查最近发出了多少个请求。

scope :in_the_last_month, -> { where('created_at > ?', Date.today - 30.days) }
# or more elegantly
scope :in_the_last_month, -> { where(created_at: 30.days.ago..) }

此代码非常简单，您实际上不需要将其变成方法，只需控制器中的代码就可以了。

if CompanyIntroduction.in_the_last_month.count >= 3
  # give some error
else
  # continue
end

用户每30天可以发送3个请求。 Ruby on Rails

1 个答案: