Question

对于会员，嵌套会员邀请，我有一个嵌套表格如下：

<%= form_for [@user, @car, @membership] do |f| %>
  <%= f.hidden_field :car_id, class: "form-control", value: @car.id %>
  <%= f.fields_for :membership_invites do |builder| %>
      <% @code = Digest::SHA1.hexdigest([Time.now, rand].join) %>
      <%= builder.hidden_field :car_id, class: "form-control", value: @car.id %>
      <%= builder.hidden_field :invite_code, class: "form-control", value: @code %>
      <div class="form-group"> 
          <%= builder.label :user_email %>
          <%= builder.text_field :user_email, class: "form-control" %>
      </div>
  <% end %>
  <%= f.submit "Add Member", class: "btn btn-large btn-primary" %>
<% end %>

它运行正常并且一切都已保存，但是，我在 MembershipInvite 控制器（创建方法）中没有调用的电子邮件中有传递调用。如果表单中的某些内容属实，我只想调用此传递方法。

如何访问成员资格控制器上的嵌套表单属性（因为此create methid似乎正在调用）。它是@ membership.membership_invites_attributes吗？

会员控制

def create

    @membership = Membership.create(membership_params)

    #I WOULD LIVE TO SEND AN EMAIL HERE DEPENDING ON FIELDS FILLED IN FROM THE MEMBERSHIP FORM AND NESTED MEMBERSHIP INVITES FORM. 

    if @membership.save
        redirect_to current_user, notice: "Membership created"
    else
        render 'new'
    end
end

Answer 1

我实际上已经设法在我的模型中使用after_create方法执行此操作。如果有人知道更好的方法，请告诉我！？

<强>模型 after_create：send_invite

def send_invite

    @user = User.where(email: self.user_email).first

    unless @user.blank?
        @membership = Membership.find(self.membership_id)
        @membership.update_attributes(:user_id => @user.id)
    else
        InviteMailer.new_invite(self).deliver
    end
end

Rails以嵌套的属性形式访问控制器操作

1 个答案: