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Manipulate the sequence of tags in xml with xsl

Posted on 2015-02-21
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Last Modified: 2015-02-25
Hi,

i need to change the sequence of the xml tag , below is the current  xml

<Summary>
    <Category type="CA">
    <Description>You no longer qualify for CA</Description>
    </Category>
    <Category type="FF">
      <Description>You no longer qualify for FF</Description>
    </Category>
    <Category type="MM">
     <Description>You no longer qualify </Description>
    </Category>
    <Category type="CC">
   <Description>Based on information you provided </Description>
    </Category>
  </Summary>

i always need FF first, then CA, CC and MM in the order.. is there a way to do this in xslt.

<Summary>
    <Category type="FF">
      <Description>You no longer qualify for FF</Description>
    </Category>
    <Category type="CA">
    <Description>You no longer qualify for CA</Description>
    </Category>
   <Category type="CC">
   <Description>Based on information you provided </Description>
    </Category>
    <Category type="MM">
     <Description>You no longer qualify </Description>
    </Category>
 
  </Summary>
 

Thank you
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Question by:sri1209
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6 Comments
 
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Expert Comment

by:Geert Bormans
ID: 40623045
The brute force approach in XSLT1

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">
    
    <xsl:strip-space elements="*"/>
    <xsl:output indent="yes"/>
      
    <xsl:template match="node()">
        <xsl:copy>
            <xsl:copy-of select="@*"/>
            <xsl:apply-templates select="node()"/>
        </xsl:copy>
    </xsl:template>
    
    <xsl:template match="Summary">
        <xsl:copy>
            <xsl:copy-of select="@*"/>
            <xsl:apply-templates select="Category[@type = 'FF']"/>
            <xsl:apply-templates select="Category[@type = 'CA']"/>
            <xsl:apply-templates select="Category[@type = 'CC']"/>
            <xsl:apply-templates select="Category[@type = 'MM']"/>
        </xsl:copy>
    </xsl:template>
    
</xsl:stylesheet>

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LVL 60

Expert Comment

by:Geert Bormans
ID: 40623063
A more refined technique in XSLT1 is described here
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/XML/Q_21571844.html
I like the elegance of my approach there, but it requires the document() function to be switched on
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LVL 60

Expert Comment

by:Geert Bormans
ID: 40623066
If you are using XSLT2, it is easy, just make a sequence of the four terms and use index-of or a function in the @select of the sort
Let me know if you like a XSLT2 solution
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LVL 60

Accepted Solution

by:
Geert Bormans earned 2000 total points
ID: 40623068
anyway, just for the completeness of the answer

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="2.0">
    
    <xsl:strip-space elements="*"/>
    <xsl:output indent="yes"/>
      
    <xsl:template match="node()">
        <xsl:copy>
            <xsl:copy-of select="@*"/>
            <xsl:apply-templates select="node()"/>
        </xsl:copy>
    </xsl:template>
    
    <xsl:template match="Summary">
        <xsl:copy>
            <xsl:copy-of select="@*"/>
            <xsl:apply-templates select="Category">
                <xsl:sort select="index-of(('FF', 'CA', 'CC', 'MM'), @type)" data-type="number"/>
            </xsl:apply-templates>
        </xsl:copy>
    </xsl:template>
    
</xsl:stylesheet>

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Author Comment

by:sri1209
ID: 40631506
Hi Geert,

Sorry for the delay in replying back, you are awesome, works perfectly.
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Expert Comment

by:Geert Bormans
ID: 40631674
welcome
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