Ruby HAML与迭代XML

Ruby HAML与迭代XML,ruby,sinatra,haml,Ruby,Sinatra,Haml,我正在尝试生成一个XML文件。我有一些数字重复的标签,如下所示: # necessary XML namespace and metadata stuff... %img{:src => "Foo002.jpg#full"}/ %img{:src => "Foo003.jpg#full"}/ %img{:src => "Foo004.jpg#full"}/ %img{:src => "Foo005.jpg#full"}/ # and so on for a few hundred lines

我正在尝试生成一个XML文件。我有一些数字重复的标签,如下所示:

# necessary XML namespace and metadata stuff... %img{:src => "Foo002.jpg#full"}/ %img{:src => "Foo003.jpg#full"}/ %img{:src => "Foo004.jpg#full"}/ %img{:src => "Foo005.jpg#full"}/ # and so on for a few hundred lines %img{:src => "Foo435.jpg#full"}/ #必要的XML命名空间和元数据内容。。。 %img{:src=>“fooo002.jpg#full”}/ %img{:src=>“Foo003.jpg#full”}/ %img{:src=>“Foo004.jpg#full”}/ %img{:src=>“Foo005.jpg#full”}/ #以此类推几百行 %img{:src=>“Foo435.jpg#full”}/ 我得到了“Foo”和结束数字(本例中为435)作为参数。我该怎么告诉哈姆来做这件事

谢谢

编辑:忘了提一下,前导零很重要

证明:

phrogz$ haml
!!! XML
%root
  - 2.upto(10) do |n|
    %img{ src:"Foo%03d.jpg" % n }
^D

<?xml version='1.0' encoding='utf-8' ?>
<root>
  <img src='Foo002.jpg' />
  <img src='Foo003.jpg' />
  <img src='Foo004.jpg' />
  <img src='Foo005.jpg' />
  <img src='Foo006.jpg' />
  <img src='Foo007.jpg' />
  <img src='Foo008.jpg' />
  <img src='Foo009.jpg' />
  <img src='Foo010.jpg' />
</root>
phrogz$haml
!!! XML
%根
-2.高达(10)do | n|
%img{src:“Foo%03d.jpg”%n}
^D
- (2..435).each do |n|
  %img{:src => "Foo%03d.jpg#full" % n}
phrogz$ haml
!!! XML
%root
  - 2.upto(10) do |n|
    %img{ src:"Foo%03d.jpg" % n }
^D

<?xml version='1.0' encoding='utf-8' ?>
<root>
  <img src='Foo002.jpg' />
  <img src='Foo003.jpg' />
  <img src='Foo004.jpg' />
  <img src='Foo005.jpg' />
  <img src='Foo006.jpg' />
  <img src='Foo007.jpg' />
  <img src='Foo008.jpg' />
  <img src='Foo009.jpg' />
  <img src='Foo010.jpg' />
</root>