Sorting XML file by attribute

Question

I have a following XML code:

<Group>
    <GElement code="x">
        <Group>
            <GElement code="x">
                <fname>a</fname>
                <lname>b</lname>
            </GElement>
            <GElement code ="f">
                <fname>fa</fname>
            </GElement>
        </Group>
    </GElement>
    <GElement code ="f">
    </GElement>
</Group>

I would like to have the output sorted by "code" like:

<Group>
    <GElement code ="f">
    </GElement>
    <GElement code="x">
        <Group>
            <GElement code ="f">
                <fname>fa</fname>
            </GElement>
            <GElement code="x">
                <fname>a</fname>
                <lname>b</lname>
            </GElement>
        </Group>
    </GElement>
</Group>

The depth of the tree can be endless i.e. the GElement can have another Group and so on.

Any ideas?

I suspect there's an XSL solution for this but I'm not posting as an answer because I'm not entirely certain. — FrustratedWithFormsDesigner
– FrustratedWithFormsDesigner, Commented May 6, 2010 at 17:00
The minute you start fretting about how your XML looks is the moment you realize you have too much time on your hands. — user1228
– user1228, Commented May 6, 2010 at 17:11
@Will: This may be true for indentation, but document order carries meaning in XML, so why not implement something that relies on document order as well? — Tomalak
– Tomalak, Commented May 6, 2010 at 17:15

Tomalak · Accepted Answer · 2010-05-06 17:53:38Z

5

Use XslCompiledTransform (see MSDN) to apply this styleshet to your XML document:

<xsl:stylesheet
  version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
>
  <!-- the identity template copies everything verbatim -->
  <xsl:template match="node()|@*">
    <xsl:copy>
      <xsl:apply-templates select="node()|@*" />
    </xsl:copy>
  </xsl:template>

  <!-- special template for <Group> that sorts its children -->
  <xsl:template match="Group">
    <xsl:copy>
      <xsl:copy-of select="@*" /> <!-- copy attributes, if any -->
      <xsl:apply-templates select="GElement">
        <xsl:sort select="@code" />
      </xsl:apply-templates>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>

XML tree nesting depth can be arbitrary.

edited May 6, 2010 at 17:53

answered May 6, 2010 at 17:02

Tomalak

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6 Comments

FrustratedWithFormsDesigner Over a year ago

...yeah, something like this, was what I was thinking.

Daniel Haley Over a year ago

@tomalak, your first xsl:template has an xsl:apply-templates end tag

Daniel Haley Over a year ago

also, is <xsl:copy-of select="@*" /> necessary? is there any difference in the output if you remove it?

Tomalak Over a year ago

@DevNull: You're right. Corrected. No, there is no difference in the output with that particular input. But if there are attributes defined in the actual XML, they'll be copied. The <xsl:copy-of select="@*" /> does not hurt or take away performance, but it makes the solution correct.

Tomalak Over a year ago

@DevNull, thinking about it, to be really correct, the solution would also need an <xsl:copy-of select="node()[not(self::GElement])" />, but I estimated that contained elements other than <GElement> were unlikely. This estimate could be wrong, so I add it as a comment. :)

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