Replace an attribute with another attribute

georgemeng · June 9th, 2008, 07:13 PM

I have an XML at the end of this post.
in this XML: I have
a variable: <variable fiscalyear="2008"/>
an attribute in tableheader element: "Actual PY YTD"

I want to create an XSLT to replace "Actual CY YTD" with "Actual 2008 YTD", is that possible?

Any suggestion is greatly appreciated. I am new to xslt, I know it is powerful, just have no clue how to do it.

Thanks very much!

<?xml version="1.0" encoding="UTF-8"?>
<result>
    <variables><variable fiscalyear="2008"/></variables>
    <filters><filter name="" type="REPORT FACTS" value="User id: ABCDE"/><filter name="" type="REPORT FACTS" value="09.06.2008"/><filter name="Year" type="FILTER SETTINGS" value="2008"/><filter name="Month" type="FILTER SETTINGS" value="3"/></filters>
    <tableheaders><tableheader header01="Structure" header02="Actual PY YTD" header03="Actual CY YTD" header04="Budget CY YTD" join="1"/></tableheaders>
    <tabledata><row data01="PUR" data02="261.03479298" data03="249.45470803" data04="233.43961195" join="1"/><row data01="PCS" data02="168.31675563" data03="164.34726647" data04="167.99465816" join="1"/><row data01="CAS" data02="97.75732698" data03="99.46597209" data04="97.67908706" join="1"/><row data01="TPU" data02="7.29061817" data03="6.78124859" data04="6.95937657" join="1"/><row data01="IBC (NEW)" data02="219.55993426" data03="215.04789106" data04="220.54434795" join="1"/><row data01="Corporate" data02="3.89445554" data03="6.24318283" data04="7.47705739" join="1"/><row data01="Admin" data02="41.45681525" data03="39.32716022" data04="48.14665681" join="1"/><row data01="Other SU" data02="107.85191851" data03="87.38876944" data04="149.55014253" join="1"/></tabledata>
    <chartheaders><chartheader header01="PUR" header02="PCS" header03="CAS" header04="TPU" header05="IBC (NEW)" header06="Corporate" header07="Admin" header08="Other SU" join="1"/></chartheaders>
    <chartdata><row data01="Actual PY YTD" data02="261.03479298" data03="168.31675563" data04="97.75732698" data05="7.29061817" data06="219.55993426" data07="3.89445554" data08="41.45681525" data09="107.85191851" join="1"/><row data01="Actual CY YTD" data02="249.45470803" data03="164.34726647" data04="99.46597209" data05="6.78124859" data06="215.04789106" data07="6.24318283" data08="39.32716022" data09="87.38876944" join="1"/><row data01="Budget CY YTD" data02="233.43961195" data03="167.99465816" data04="97.67908706" data05="6.95937657" data06="220.54434795" data07="7.47705739" data08="48.14665681" data09="149.55014253" join="1"/></chartdata>
</result>

George Meng

georgemeng · June 9th, 2008, 07:28 PM

OK, after reading the guide to post a new question, I think it will be good for me to change the question a little bit so it is easier for people to understand.

Sorry guys.

Here is the new question:

I have an xml:

<?xml version="1.0"?>
<result>
    <variables>
        <variable fiscalyear="2008"/>
    </variables>
    <tableheaders>
        <tableheader header02="Actual PY YTD" header03="Actual CY YTD" header04="Budget CY YTD"/>
    </tableheaders>
    <chartdata>
        <row data01="Actual PY YTD" data02="261.03479298" />
        <row data01="Actual CY YTD" data02="249.45470803" />
        <row data01="Budget CY YTD" data02="233.43961195" />
    </chartdata>
</result>

I want to create an XSLT to convert "Actual CY YTD" to "Actual 2008 YTD", where 2008 is in element variable as an attribute.

Here is the XSLT, you can see that currently I just copy the XML, no convert yet.

My question is how can I do that in XSLT: replace "CY" with "2008"?

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

    <xsl:template match="result">
        <result>
            <xsl:apply-templates/>
        </result>
    </xsl:template>

    <xsl:template match="/result/variables">
        <xsl:copy>
            <xsl:copy-of select="variable"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="/result/tableheaders">
        <xsl:copy>
            <xsl:copy-of select="tableheader"/>
        </xsl:copy>
    </xsl:template>

    <xsl:template match="/result/chartdata">
        <xsl:copy>
            <xsl:copy-of select="row"/>
        </xsl:copy>
    </xsl:template>

</xsl:stylesheet>

Thanks very much!

George Meng

joefawcett · June 10th, 2008, 02:27 AM

Somebody read the sticky post!
This is not my best ever effort but should start you off. It would be easier in XSLT 2.0 which allows user defined functions, in 1.0 you need to use named templates:

Code:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:output method="xml"/>
  <xsl:template match="result">
    <result>
      <xsl:apply-templates/>
    </result>
  </xsl:template>
  <xsl:template match="/result/variables">
    <xsl:copy>
      <xsl:copy-of select="variable"/>
    </xsl:copy>
  </xsl:template>
  <xsl:template match="/result/tableheaders">
    <xsl:copy>
      <xsl:copy-of select="tableheader"/>
    </xsl:copy>
  </xsl:template>
  <xsl:template match="/result/chartdata">
    <xsl:copy>
      <xsl:apply-templates select="row"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="row">
    <xsl:variable name="data01Replacement">
      <xsl:choose>
        <xsl:when test="contains(@data01, ' CY ')">
          <xsl:call-template name="replaceCY">
            <xsl:with-param name="cyText" select="@data01"/>
            <xsl:with-param name="year" select="/*/variables/variable/@fiscalyear"/>
          </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
          <xsl:value-of select="@data01"/>
        </xsl:otherwise>
      </xsl:choose>
    </xsl:variable>
    <row data01="{$data01Replacement}" data02="@data02"/>
  </xsl:template>

  <xsl:template name="replaceCY">
    <xsl:param name="cyText"/>
    <xsl:param name="year"/>
    <xsl:value-of select="substring-before($cyText, 'CY')"/>
    <xsl:value-of select="$year"/>
    <xsl:value-of select="substring-after($cyText, 'CY')"/>
  </xsl:template>
</xsl:stylesheet>

--

Joe (Microsoft MVP - XML)

mhkay · June 10th, 2008, 02:55 AM

I would do this using the "modified identity template" coding pattern.

First write a template (the "identity template" that copies everything unchanged, recursively:

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

Then override it with rules for things you want to change:

<xsl:template match="@*[.='Actual CY YTD']">
  <xsl:attribute name="{name()}">Actual 2008 YTD</xsl:attribute>
</xsl:template>

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

georgemeng · June 10th, 2008, 10:51 AM

Thanks Joe, that is what I need.

Quote:

quote:Originally posted by joefawcett
Somebody read the sticky post!
This is not my best ever effort but should start you off. It would be easier in XSLT 2.0 which allows user defined functions, in 1.0 you need to use named templates:

Code:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:output method="xml"/>
  <xsl:template match="result">
    <result>
      <xsl:apply-templates/>
    </result>
  </xsl:template>
  <xsl:template match="/result/variables">
    <xsl:copy>
      <xsl:copy-of select="variable"/>
    </xsl:copy>
  </xsl:template>
  <xsl:template match="/result/tableheaders">
    <xsl:copy>
      <xsl:copy-of select="tableheader"/>
    </xsl:copy>
  </xsl:template>
  <xsl:template match="/result/chartdata">
    <xsl:copy>
      <xsl:apply-templates select="row"/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="row">
    <xsl:variable name="data01Replacement">
      <xsl:choose>
        <xsl:when test="contains(@data01, ' CY ')">
          <xsl:call-template name="replaceCY">
            <xsl:with-param name="cyText" select="@data01"/>
            <xsl:with-param name="year" select="/*/variables/variable/@fiscalyear"/>
          </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
          <xsl:value-of select="@data01"/>
        </xsl:otherwise>
      </xsl:choose>
    </xsl:variable>
    <row data01="{$data01Replacement}" data02="@data02"/>
  </xsl:template>

  <xsl:template name="replaceCY">
    <xsl:param name="cyText"/>
    <xsl:param name="year"/>
    <xsl:value-of select="substring-before($cyText, 'CY')"/>
    <xsl:value-of select="$year"/>
    <xsl:value-of select="substring-after($cyText, 'CY')"/>
  </xsl:template>
</xsl:stylesheet>

--

Joe (Microsoft MVP - XML)

George Meng

georgemeng · June 10th, 2008, 10:54 AM

Hi Jay,

I like you code, just one more question: Is there a way not hard code "2008" in your code?

<xsl:attribute name="{name()}">Actual 2008 YTD</xsl:attribute>

Thanks a lot!

Quote:

quote:Originally posted by mhkay
I would do this using the "modified identity template" coding pattern.

First write a template (the "identity template" that copies everything unchanged, recursively:

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

Then override it with rules for things you want to change:

<xsl:template match="@*[.='Actual CY YTD']">
<xsl:attribute name="{name()}">Actual 2008 YTD</xsl:attribute>
</xsl:template>

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

George Meng

samjudson · June 10th, 2008, 10:56 AM

Code:

<xsl:attribute name="{name()}"Actual <xsl:value-of select="/result/variables/variable/@fiscalyear"/> YTD</xsl:attribute>

/- Sam Judson : Wrox Technical Editor -/

georgemeng · June 10th, 2008, 10:58 AM

Hi Michael,

I am so sorry, I type the wrong name.

Best Regards,

Quote:

quote:Originally posted by georgemeng
Hi Jay,

I like you code, just one more question: Is there a way not hard code "2008" in your code?
 
<xsl:attribute name="{name()}">Actual 2008 YTD</xsl:attribute>

Thanks a lot!

Quote:

quote:Originally posted by mhkay
I would do this using the "modified identity template" coding pattern.

First write a template (the "identity template" that copies everything unchanged, recursively:

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

Then override it with rules for things you want to change:

<xsl:template match="@*[.='Actual CY YTD']">
<xsl:attribute name="{name()}">Actual 2008 YTD</xsl:attribute>
</xsl:template>

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

George Meng

georgemeng · June 10th, 2008, 11:04 AM

Thanks you so much Sam, love you guys!

Best Regards!

Quote:

quote:Originally posted by samjudson

Code:

<xsl:attribute name="{name()}"Actual <xsl:value-of select="/result/variables/variable/@fiscalyear"/> YTD</xsl:attribute>

/- Sam Judson : Wrox Technical Editor -/

George Meng