如何從rails中的多個模型中選擇數據

Question

我有三個模型 project_manager、project_director 和 human_resource，每個模型都有一個狀態布爾字段，如果這三個模型的布爾值為真，我如何在 rails 中打印一些東西。 目前我正在通過這樣做訪問模型中的數據-

            <% if project_site.project_managers.empty? %>
              <td class="pending fi-eye"><%= " Pending" %></td>
            <% else %>
              <% project_site.project_managers.each do |project_manager| %>
                <% if project_manager.status == false %>
                  <td class="rejected fi-x"><%= ' Rejected' %></td>
                <% elsif project_manager.status == true %>
                  <td class="approved fi-check"><%= " Approved" %></td>
                <% end %>
              <% end %>
            <% end %>

            <% if project_site.project_directors.empty? %>
              <td class="pending fi-eye"><%= " Pending" %></td>
            <% else %>
              <% project_site.project_directors.each do |project_director| %>
                <% if project_director.status == false %>
                  <td class="rejected fi-x"><%= ' Rejected' %></td>
                <% elsif project_director.status == true %>
                  <td class="approved fi-check"><%= " Approved" %></td>
                <% end %>
              <% end %>
            <% end %>

            <% if project_site.human_resources.empty? %>
              <td class="pending fi-eye"><%= " Pending" %></td>
            <% else %>
              <% project_site.human_resources.each do |human_resource| %>
                <% if human_resource.status == false %>
                  <td class="rejected fi-x"><%= ' Rejected' %></td>
                <% elsif human_resource.status == true %>
                  <td class="approved fi-check"><%= " Approved" %></td>
                <% end %>
              <% end %>
            <% end %>

如果這三個模型狀態值都為真，我想打印已批准我怎么能在 Rails 中做到這一點？

Answer 1

創建一個輔助方法並輸入以下代碼：

def check_resource_status(project_site, resources)
 statuses = project_site.send(resources.to_sym).pluck(:status)
 statuses.all? ? true : false
end

def status_container(status)
 content_tag :div, class: ['sample'] do
  status_label = status ? 'Approved' : 'Rejected'
  default_class = status ? 'fi-check' : 'fi-x'
  status_class = [default_class, status_label.downcase]
  concat content_tag(:label, status_label, class: status_class)
 end
end

從您的視圖文件：

status_container(check_resource_status(project_site, 'human_resources'))
status_container(check_resource_status(project_site, 'project_directors'))
status_container(check_resource_status(project_site, 'project_managers'))

Answer 2

如果我正確理解了你的問題，你所需要的就是這樣的：

<% if human_resource.status? %>
  <td class="approved fi-check"><%= ' Approved' %></td>
<% else %>
  <td class="rejected fi-x"><%= ' Rejected' %></td>
<% end %>