How to counting nodes by name in xml file by xslt?

Question

I'm new with xsl . I am having below xml:

    <record>
<fruit>Apples</fruit>
<fruit>Oranges</fruit>
<fruit>Bananas</fruit>
<fruit>Plums</fruit>
<vegetable>Carrots</vegetable>
<vegetable>Peas</vegetable>
<candy>Snickers</candy>

I want to use key function and have an output file:

     <record> 1
--<fruit> 4
--<vegetable> 2
--<candy> 1

Any solutions?

Are those hyphens part of the expected output? Do you really want output with unclosed XML tags? — JLRishe
– JLRishe, Commented Jan 12, 2013 at 6:01

Dimitre Novatchev · Accepted Answer · 2013-01-12 04:11:31Z

2

As easy as this:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output method="text"/>
 <xsl:key name="kElemByName" match="*" use="name()"/>

 <xsl:template match=
  "*[not(generate-id()=generate-id(key('kElemByName',name())[1]))]"/>

 <xsl:template match="*">
  <xsl:value-of select=
  "concat('&lt;',name(),'> ', count(key('kElemByName',name())),'&#xA;')"/>
  <xsl:apply-templates select="*"/>
 </xsl:template>
</xsl:stylesheet>

When this transformation is applied on the provided XML document:

<record>
    <fruit>Apples</fruit>
    <fruit>Oranges</fruit>
    <fruit>Bananas</fruit>
    <fruit>Plums</fruit>
    <vegetable>Carrots</vegetable>
    <vegetable>Peas</vegetable>
    <candy>Snickers</candy>
</record>

the wanted, correct result is produced:

<record> 1
<fruit> 4
<vegetable> 2
<candy> 1

answered Jan 12, 2013 at 4:11

Dimitre Novatchev

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1 Comment

Do Thanh Tung Over a year ago

Can you show me the way have result:/n <record> 1 --<fruit> 4 --<veget> 2 --<candi> 1

JLRishe · Accepted Answer · 2013-01-12 15:02:25Z

1

To include the hyphens (I will use Dimitre's answer as a base, so please give him the credit he is due), you can do this:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="text"/>
  <xsl:key name="kElemByName" match="*" use="name()"/>

  <xsl:template match=
  "*[not(generate-id()=generate-id(key('kElemByName',name())[1]))]">
    <xsl:apply-templates select="*" />
  </xsl:template>

  <xsl:template match="*">

    <xsl:apply-templates select="ancestor::*" mode="hyphens" />

    <xsl:value-of select=
       "concat('&lt;',name(),'> ', count(key('kElemByName',name())),'&#xA;')"/>
    <xsl:apply-templates select="*"/>
  </xsl:template>

  <xsl:template match="*" mode="hyphens">
    <xsl:text>--</xsl:text>
  </xsl:template>
</xsl:stylesheet>

With the original input, this produces:

<record> 1
--<fruit> 4
--<vegetable> 2
--<candy> 1

With a more deeply nested input,

<record>
  <fruit>
    Apples
  </fruit>
  <fruit>
    <citrus>Grapefruits</citrus>
    <citrus>Oranges</citrus>
    <citrus>Lemons</citrus>
  </fruit>
  <fruit>Bananas</fruit>
  <fruit>
    <pitted>Plums</pitted>
    <pitted>Apricots</pitted>
    <pitted>Peaches</pitted>
  </fruit>
  <vegetable>Carrots</vegetable>
  <vegetable>Peas</vegetable>
  <candy>
    <chocolate>Snickers</chocolate>
    <chocolate>Milky Way</chocolate>
    <chocolate>Hersheys</chocolate>
  </candy>
  <candy>
    <hard>Lozenges</hard>
    <hard>Lollipops</hard>
  </candy>
</record>

You get:

<record> 1
--<fruit> 4
----<citrus> 3
----<pitted> 3
--<vegetable> 2
--<candy> 2
----<chocolate> 3
----<hard> 2

How's that?

answered Jan 12, 2013 at 15:02

JLRishe

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2 Comments

Ian Roberts Over a year ago

+1 for the <xsl:apply-templates select="ancestor::*" mode="hyphens" /> - very elegant, and an idiom I will have to remember for future reference.

JLRishe Over a year ago

Thanks @IanRoberts! Here's another case where I used a similar technique: stackoverflow.com/a/14274683/1945651 In this post here, a <xsl:for-each select="ancestor::*"><xsl:text>--<xsl:text></xsl:for-each> could've been just as good, but the other thread actually makes use of the template in more than one place.

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