Increment a XML attribute with XSLT?

Question

<Menu>
  <Food>
    <Item>
      <Name>Waffles</Name>
      <Price>$9</Price>
      <Description>Try them with syrup *Additional Charge*</Description>
    </Item>
    <Item name="Pop-Tarts" price="$40" description="yummy"></Item>
  <Food>
<Menu>

I used the following XSLT to break up the second food item into three elements

<xsl:template match="/">
  <xsl:apply-templates select="Menu/Food" />
</xsl:template>

<xsl:template match="@* | node()">
  <xsl:copy>
    <xsl:apply-templates select="@* | node()" />
  </xsl:copy>
</xsl:template>

<xsl:template match="Item">
  <xsl:element name="{name()}">
    <!--This is what Im trying to fix-->
    <xsl:attribute name="id"></xsl:attribute>

    <xsl:for-each select="@*">
      <xsl:element name="{name()}">
        <xsl:value-of select="."/>
      </xsl:element>
    </xsl:for-each>
    <xsl:apply-templates select="* | text()"/>
  </xsl:element>
</xsl:template>

In the last template, how would I take the 'id' attribute for each 'Item' element and increment it for each new item?

This is what my current output looks like

<Food>
  <Item id="">
    <Name>Waffles</Name>
    <Price>$9</Price>
    <Description>Try them with syrup *Additional Charge*</Description>
  </Item>
  <Item id="">
    <name>Pop-Tarts</name>
    <price>$40</price>
    <description>yummy</description>
  </Item>
</Food>

kjhughes · Accepted Answer · 2020-05-22 21:27:25Z

2

How about:

XSLT 1.0

<xsl:stylesheet version="1.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

<xsl:template match="/Menu">
    <Food>
        <xsl:for-each select="Food/Item">
            <Item id="{position()}">
                <xsl:for-each select="@*">
                    <xsl:element name="{name()}">
                        <xsl:value-of select="."/>
                    </xsl:element>
                </xsl:for-each>
                <xsl:copy-of select="*"/>
            </Item>
        </xsl:for-each>
    </Food>
</xsl:template>

</xsl:stylesheet>

Note that XML is case-sensitive: Name is not the same as name.

edited May 22, 2020 at 21:27

kjhughes

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answered May 22, 2020 at 21:16

michael.hor257k

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Comments

Austin Garrison · Accepted Answer · 2020-05-26 19:29:22Z

0

@michael.hor257k This code ends up not displaying anything in my output.

However using this part of your answer in my code got what I needed.

<xsl:template match="Food/Item">
  <Item id="{position()}">
    .....
  </Item>
</xsl:template>

answered May 26, 2020 at 19:29

Austin Garrison

194 bronze badges

1 Comment

michael.hor257k Over a year ago

You can see my answer working (with input corrected to be well-formed XML) here: xsltfiddle.liberty-development.net/jxDiMCh.

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