Jim Andersen wrote:
> I have this xml-file. I want to sort it, and create a
> new xml-file. The result should be identical to the
> input except that it's sorted.

Since you failed to mention what you're using for
processing, I assume it's some XSLT 1.0-compliant
processor.

> <?xml version="1.0" encoding="UTF-8"?>
> <levelone>
> <child ID="1" sort="5"><name>Paul</name></child>
> <child ID="2" sort="1"><name>Adam</name></child>
> <child ID="3" sort="2"><name>Will</name></child>
> <root>

That's not an XML file. It's not well-formed.

> I've been searching for hours, but all the examples I
> have seen creates an HTML file.

That's mighty strange, especially since XML is the
default output method. All the XSLT tutorials I remember
covered sorting, identity transformation and controlling
your output.

Presuming you meant something like:

<levelone>
<child ID="1" sort="5"><name>Paul</name></child>
<child ID="2" sort="1"><name>Adam</name></child>
<child ID="3" sort="2"><name>Will</name></child>
</levelone>

The following XSLT should do the trick:

<xsl:stylesheet version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="node()|@*">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
  </xsl:template>
  <xsl:template
    match="*[@sort][not(preceding-sibling::*[@sort])]">
    <xsl:apply-templates select="../*[@sort]" mode="copy">
      <xsl:sort select="@sort"/>
    </xsl:apply-templates>
  </xsl:template>
  <xsl:template
    match="*[@sort][preceding-sibling::*[@sort]]"/>
  <xsl:template match="node()|@*" mode="copy">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*"/>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

-- 
Pavel Lepin