How to unencode escaped XML with xQuery

I have a variable in xQuery of type xs:string with the value of an encoded HTML snippet (the content of a twitter tweet). It looks like this:

Headlines-Today • AP sources: <b>Obama</b> pick for Justice post withdraws : News - Rest Of World - <a href="http://shar.es/mqMAG">http://shar.es/mqMAG</a>

When I try to write this out in an HTML block, I need the string to开发者_JAVA百科 be unescaped so that the HTML snippet will be interpreted by the browser. Instead the string is getting written out as is and the browser is rendering it as just text (so you see <a href="blah.... ). Here's how I'm writing out this string:

{$entry/atom:content/text()}

How can I have the escaped characters unencoded so it writes < rather tha &lt; ?

I've tried to do a replacelike this but it always replaces the &lt; with &lt; !

fn:replace($s, "&lt;", "<")

In MarkLogic you can use the below query:

let $d := '<a>&lt;c&gt;asdf&lt;/c&gt;</a>' 

return xdmp:unquote ($d)

in eXist, use util:parse():

util:parse(concat("<top>","&lt;c&gt;asdf&lt;/c&gt;",</top>")‌​)

Depends on which XQuery processor you are using... The easiest way is to be using a processor that has an extension that handles this for you. For instance, with Saxon and the following XML:

<a>&lt;c&gt;asdf&lt;/c&gt;</a>

You can write an XQuery that uses the saxon:parse() function to do what you want:

declare namespace saxon = "http://saxon.sf.net/";

<a>{
  saxon:parse(doc('test.xml')/a)
}</a>

The result from that is:

<a>
  <c>asdf</c>
</a>

I think most(?) XQuery processors will have an extension to do this for you. Hope that helps.

since XQuery 3 you can use the parse-xml() or parse-xml-fragment() functions. See also:

• https://www.w3.org/TR/xpath-functions-3/#func-parse-xml
• https://www.w3.org/TR/xpath-functions-3/#func-parse-xml-fragment