2015-04-01 65 views
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<recipes> 
    <recipe name="MPEG4w480f20h360AAC" mimeType="video/mp4" width="480" height="360" /> 
    <recipe name="MP3" mimeType="audio/mpeg" /> 
    <recipe name="j24" mimeType="image/jpeg" width="24" height="24" /> 
    <recipe name="j64" mimeType="image/jpeg" width="64" height="64" /> 
    <recipe name="j128" mimeType="image/jpeg" width="128" height="128" /> 
    <recipe name="j88" mimeType="image/jpeg" height="88" /> 
    <recipe name="j150" mimeType="image/jpeg" height="150" /> 
    <recipe name="hqPivot" mimeType="image/jpeg" width="600" /> 
    </recipes> 

我在写Ruby脚本来验证XML结构。如何使用ruby解析XML以获得重复的标记?

assert(!XPath.match(xml, "recipes/recipe[@name]").empty?, "Structure of xml incorrect") 

如何才能一一验证?

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,我没有得到你的问题? – 2015-04-01 09:39:27

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如何验证下的所有[@names]? – Zaf 2015-04-01 11:14:47

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'//食谱/食谱[@name]'会给所有有'@name的食谱' – 2015-04-01 11:29:36

回答

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使用引入nokogiri:

require 'nokogiri' 

f = File.open("test.xml") 
doc = Nokogiri::XML(f) 

tags = [] 

doc.xpath("//recipe/@name").each do |elem| # get all name of all recipes 
    tags << elem.value 
end 

p tags # show all tags 
p tags == tags.uniq # are all tags unique? 

f.close 

此输出(使用您的样品):

["MPEG4w480f20h360AAC", "MP3", "j24", "j64", "j128", "j88", "j150", "hqPivot"] 
true 
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非常感谢你Martin。 – Zaf 2015-04-01 11:56:17