Transform XML using XSLT

Question

I want to transform the XML using XSLT Here is my sample XML

<EmployeList>
  <EMpDetails>
    <Name>Kiran</Name>
    <ID>ID001</ID>
    <city>Hyderabad</city>
    <Country>India</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>Sunny</Name>
    <ID>ID002</ID>
    <city>Banglore</city>
    <Country>INDIA</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>John</Name>
    <ID>ID001</ID>
    <city>TEXAS</city>
    <Country>US</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>Raj</Name>
    <ID>ID006</ID>
    <city>Dallas</city>
    <Country>US</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>Nag</Name>
    <ID>ID007</ID>
    <city>ITALY</city>
    <Country>Rome</Country>
  </EMpDetails>
</EmployeList>

Required output using XSLT

<EmployeList>
  <EMpDetails>
    <Name>Kiran</Name>
    <ID>ID001</ID>
    <city>Hyderabad</city>
    <Country>India</Country>
  </EMpDetails>
  <EMpDetails>
    <Name>Sunny</Name>
    <ID>ID002</ID>
    <city>Banglore</city>
    <Country>INDIA</Country>
  </EMpDetails>   
</EmployeList>

And what is the rule? Like, remove everything that is not <Country>India</Country>? — GSerg
– GSerg, Commented Jun 12, 2011 at 17:05
Good question, +1. See my answer for two complete solutions (for XSLT 1.0 and XSLT 2.0) that are based on the most fundamental and powerful XSLT design pattern -- the overriding of the identity rule/template. Explanation is also provided. — Dimitre Novatchev
– Dimitre Novatchev, Commented Jun 12, 2011 at 18:22
Should be City=Rome and Country=ITALY and not the other way around :) — Emiliano Poggi
– Emiliano Poggi, Commented Jun 12, 2011 at 19:34

Dimitre Novatchev · Accepted Answer · 2011-06-12 18:20:12Z

1

Assuming that the question asks to output only EMpDetails whose Country child has a string (case-insensitive) value of "India":

I. This XSLT 1.0 transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

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

 <xsl:template match=
 "EMpDetails
    [not(translate(Country,
                   'ABCDEFGHIJKLMNOPQRSTUVWXYZ',
                   'abcdefghijklmnopqrstuvwxyz'
                   )
         ='india'
         )
    ]
 "/>
</xsl:stylesheet>

when applied on the provided XML document:

<EmployeList>
    <EMpDetails>
        <Name>Kiran</Name>
        <ID>ID001</ID>
        <city>Hyderabad</city>
        <Country>India</Country>
    </EMpDetails>
    <EMpDetails>
        <Name>Sunny</Name>
        <ID>ID002</ID>
        <city>Banglore</city>
        <Country>INDIA</Country>
    </EMpDetails>
    <EMpDetails>
        <Name>John</Name>
        <ID>ID001</ID>
        <city>TEXAS</city>
        <Country>US</Country>
    </EMpDetails>
    <EMpDetails>
        <Name>Raj</Name>
        <ID>ID006</ID>
        <city>Dallas</city>
        <Country>US</Country>
    </EMpDetails>
    <EMpDetails>
        <Name>Nag</Name>
        <ID>ID007</ID>
        <city>ITALY</city>
        <Country>Rome</Country>
    </EMpDetails>
</EmployeList>

produces the wanted, correct result:

<EmployeList>
   <EMpDetails>
      <Name>Kiran</Name>
      <ID>ID001</ID>
      <city>Hyderabad</city>
      <Country>India</Country>
   </EMpDetails>
   <EMpDetails>
      <Name>Sunny</Name>
      <ID>ID002</ID>
      <city>Banglore</city>
      <Country>INDIA</Country>
   </EMpDetails>
</EmployeList>

II. This XSLT 2.0 transformation:

<xsl:stylesheet version="2.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

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

 <xsl:template match=
 "EMpDetails[not(upper-case(Country)='INDIA')]"/>
</xsl:stylesheet>

again produces the same, wanted and correct result.

Explanation: Overriding the identity rule with a template matching all unwanted EMpDetails elements. This template has empty body, which effectively prevents copying any such matched elements to the output.

Remember: Using and overriding the identity rule/template is the most fundamental and most powerful XSLT design pattern.

answered Jun 12, 2011 at 18:20

Dimitre Novatchev

244k27 gold badges307 silver badges438 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Tomalak Over a year ago

If 'india' is fixed, not(translate(Country, 'INDA', 'inda') = 'india') would probably be more efficient.

kiran425 Over a year ago

HI Thanks for the reply and solution. can you provide me any docs to get more XSLT

Dimitre Novatchev Over a year ago

@kiran425: See this: stackoverflow.com/questions/339930/…

Emiliano Poggi · Accepted Answer · 2011-06-12 18:23:30Z

0

That should work as you need (assuming that you want all EMpDetails elements where Country is equal to India with case-insensitivity):

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
    <xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

    <xsl:template match="EmployeList">
        <EmployeList>
            <xsl:copy-of select="EMpDetails[lower-case(Country) = 'india']"/>
        </EmployeList>
    </xsl:template>
</xsl:stylesheet>

edited Jun 12, 2011 at 18:23

Emiliano Poggi

24.9k8 gold badges58 silver badges66 bronze badges

answered Jun 12, 2011 at 17:18

Grzegorz Szpetkowski

38.1k6 gold badges94 silver badges140 bronze badges

5 Comments

Dimitre Novatchev Over a year ago

There is a small problem in your solution -- a wrong and misleading version attribute that would lead people to believe they can use this with an XSLT 1.0 processor. Please, correct.

Dimitre Novatchev Over a year ago

Thanks to @empo for editing the answer and correcting the version. I am still not deleting my comment so that the original author confirms he has seen it and will know not to make such mistakes in the future.

Grzegorz Szpetkowski Over a year ago

@Dimitre Novatchev, @empo: Thanks for correcting. lower-case() function is available in XPath 2.0 (strange that Oxygen XML didn't complain about that..). For 1.0 version translate() function made the job.

kiran425 Over a year ago

HI Thanks for the reply and solution can you provide me the any docs on XML to XML transformation using XSLT

Grzegorz Szpetkowski Over a year ago

@kiran425: At beginning you can start with W3schools XML Tutorials (w3schools.com): (Learn XML), Learn XPath, Learn XSLT (in practice you can't write XSLT without knowing XPath). There aren't fully comprehensive, but easy to learn.

Emiliano Poggi · Accepted Answer · 2011-06-12 19:11:48Z

0

Assuming your sample input is simple as it's shown you could also go with:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output method="xml" indent="yes"/>
    <xsl:strip-space elements="*"/>

    <xsl:template match="/EmployeList">
        <xsl:copy>
            <xsl:copy-of select="EMpDetails[
                Country[
                .='India'
                or .='INDIA']
                ]"/>
        </xsl:copy>
    </xsl:template>

</xsl:stylesheet>

answered Jun 12, 2011 at 19:11

Emiliano Poggi

24.9k8 gold badges58 silver badges66 bronze badges

1 Comment

kiran425 Over a year ago

Hi,Thanks for the reply and solution can send me any link or documents to study more on XSLT

Collectives™ on Stack Overflow

Transform XML using XSLT

3 Answers 3

3 Comments

5 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

5 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related