通過日期整數

Question

我有下面的代碼循環：

<span class="bookings"> 
    <span class="col-md-2"> 
     <h4><%= time_tag(Date.today + 1.days) %></h4> 
     <% @booking.each do |b| %> 
     <% if b.date == Date.today + 1.days && b.type == "Hot Desk" %> 
      <% if b.desk == @hotdesk.code %> 
      <span class="glyphicon glyphicon-remove booked-show"></span> 
      <% end %> 
     <% end %> 
     <% end %> 
    </span> 
    </span> 

    <span class="bookings"> 
    <span class="col-md-2"> 
     <h4><%= time_tag(Date.today + 2.days) %></h4> 
     <% @booking.each do |b| %> 
     <% if b.date == Date.today + 2.days && b.type == "Hot Desk" %> 
      <% if b.desk == @hotdesk.code %> 
      <span class="glyphicon glyphicon-remove booked-show"></span> 
      <% end %> 
     <% end %> 
     <% end %> 
    </span> 
    </span> 

其中關於延伸到然而，許多天，我想顯示的預訂的可用性。我想要實現的只是寫出這些代碼塊中的一個，並讓它循環通過它說的地方Date.today + 1.days並將每個循環的值增加1個循環達20次左右。

任何想辦法做到這一點，以保存我一次又一次地寫這段代碼？

Answer 1

<span class="bookings"> 
    <% 1.upto(YOUR_COUNT.to_i).each do |day_count| %> 
    <span class="col-md-2"> 
     <h4><%= time_tag(Date.today + day_count.days) %></h4> 
     <% @booking.each do |b| %> 
     <% if b.date == Date.today + day_count.days && b.type == "Hot Desk" %> 
      <% if b.desk == @hotdesk.code %> 
      <span class="glyphicon glyphicon-remove booked-show"></span> 
      <% end %> 
     <% end %> 
     <% end %> 
    </span> 
    <% end %> 
    </span>