Parsing simple XML with Nokogiri

Question

I have the following XML:

<links>

  <item>
    <title>Title 1</title>
    <url>http://www.example.com/url-1</url>
  </item>

  <item>
   <title>Title 2</title>
   <url>http://www.example.com/url-2</url>
  </item>

  <item>
    <title>Title 3</title>
    <url>http://www.example.com/url-3</url>
  </item>

</links>

And, I would like to convert it to a HTML list:

<ul>
  <li><a href="http://www.example.com/url-1">Title 1</a></li>
  <li><a href="http://www.example.com/url-2">Title 2</a></li>
  <li><a href="http://www.example.com/url-3">Title 3</a></li>
</ul>

Currently I have this:

Controller:

require 'nokogiri'
doc = Nokogiri::XML(...)

@links = doc.xpath('//links/item').map do |i|
  {'title' => i.xpath('//title'), 'url' => i.xpath('//url')}
end

Template:

<ul>
  <% @links.each do |l| %>
    <li><a href="<%= l['url'] %>"><%= l['title'] %></a></li>
  <% end %>
</ul>

Resulting HTML:

<ul>
  <li><a href="http://www.example.com/url-1http://www.example.com/url-2http://www.example.com/url-3">Title 1Title 2Title 3</a></li>
  <li><a href="http://www.example.com/url-1http://www.example.com/url-2http://www.example.com/url-3">Title 1Title 2Title 3</a></li>
  <li><a href="http://www.example.com/url-1http://www.example.com/url-2http://www.example.com/url-3">Title 1Title 2Title 3</a></li>
</ul>

What am I doing wrong? Is there a more optimal way of doing this?

Good question, +1. See my answer for a correct solution and for a detailed explanation of the problem. — Dimitre Novatchev
– Dimitre Novatchev, Commented Oct 15, 2010 at 1:45

Dimitre Novatchev · Accepted Answer · 2010-10-15 01:45:01Z

29

Replace this:

@links = doc.xpath('//links/item').map do |i| 
  {'title' => i.xpath('//title'), 'url' => i.xpath('//url')}

with:

@links = doc.xpath('//links/item').map do |i| 
  {'title' => i.xpath('title'), 'url' => i.xpath('url')}

Explanation:

//title

and

//url

are absolute XPath expressions and they select all (respectively) title and all url elements in the XML document.

Contrast this with:

title

and

url

These are relative XPath expressions and select all (respectively) title and url children of the current node only.

answered Oct 15, 2010 at 1:45

Dimitre Novatchev

244k27 gold badges308 silver badges438 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Matchu Over a year ago

Answer revoked, +1, since I assume you actually know what you're talking about. I don't know Xpath, and just guessed xD

Dimitre Novatchev Over a year ago

@Matchu: Yes, I do know XPath and happen to be ranked #1 by rep in this tag. :) But your answer was correct -- you needn't delete it. Undelete it and I'll upvote.

Matchu Over a year ago

Thanks :) I'm usually pretty picky about just letting one answer be up when there are dupes within seconds, though, since my OCD wins out over my insatiable hunger for rep. Thanks, though! You are a gentleman and a scholar :o

Dimitre Novatchev Over a year ago

@Matchu: Please... Undelete your answer. I want to upvote it.

Matchu · Accepted Answer · 2010-10-15 01:44:47Z

6

The trouble here is that the Xpath //title searches for titles from the root of the document, and so returns all title tags. Using the Xpath title searches within the context of the given node, like you want. Ditto on url.

@links = doc.xpath('//links/item').map do |i|
  {'title' => i.xpath('title'), 'url' => i.xpath('url')}
end

answered Oct 15, 2010 at 1:44

Matchu

86.1k18 gold badges155 silver badges160 bronze badges

1 Comment

Dimitre Novatchev Over a year ago

Wow... I saw your undelete only now. Of course, the fully deserved +1. And I admired your explanation of the RegEx that decides if a number representation is composite!!!

Collectives™ on Stack Overflow

Parsing simple XML with Nokogiri

2 Answers 2

4 Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

4 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related