how to modify xml file using xslt

Question

I need to change the value of particular element of a node in a xml through xsl file Below is my xml data

<hospitals>
  <hospital>
    <department>
      <clinic>
        <cid>8</cid>
        <clinicName>clinic8</clinicName>
        <status>1</status>
      </clinic>
      <clinic>
        <cid>9</cid>
        <clinicName>clinic9</clinicName>
        <status>0</status>
      </clinic>
      <depId>3</depId>
      <departmentName>dental</departmentName>
    </department>
    <hospId>2</hospId>
    <hospitalName>appolo</hospitalName>
  </hospital>
  <hospital>
    <department>
      <clinic>
        <cid>82</cid>
        <clinicName>clinic82</clinicName>
        <status>0</status>
      </clinic>
      <clinic>
        <cid>92</cid>
        <clinicName>clinic92</clinicName>
        <status>0</status>
      </clinic>
      <depId>4</depId>
      <departmentName>mental</departmentName>
    </department>
    <hospId>2</hospId>
    <hospitalName>manipal</hospitalName>
  </hospital>
</hospitals>

for example, I need to select clinic9 based on its id ie 9 and change the status 0 to 1

I tried like this

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:param name="hospId"/>
<xsl:param name="depId" />
<xsl:param name="clinicId"/>
<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="hospitals/hospital[hospId='2']/department[depId='3']/clinic[cid='9']">
<xsl:choose>
 <xsl:when test="cid ='9'">
  <xsl:element name="status">123</xsl:element>
  </xsl:when>
</xsl:choose>
</xsl:template>
  </xsl:stylesheet>

But the value is not changing...

Daniel Haley · Accepted Answer · 2012-08-02 07:13:29Z

8

If you are trying to modify/replace a specific element, you need to match that element. For example, if you are trying to replace a specific status element, you need to match that specific element.

Modified XSLT

<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:param name="hospId" select="'2'"/>
    <xsl:param name="depId" select="'3'"/>
    <xsl:param name="clinicId" select="'9'"/>

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

    <xsl:template match="hospitals/hospital[hospId='2']/department[depId='3']/clinic[cid='9']/status">
        <status>123</status>
    </xsl:template>
</xsl:stylesheet>

XML Output

<hospitals>
   <hospital>
      <department>
         <clinic>
            <cid>8</cid>
            <clinicName>clinic8</clinicName>
            <status>1</status>
         </clinic>
         <clinic>
            <cid>9</cid>
            <clinicName>clinic9</clinicName>
            <status>123</status>
         </clinic>
         <depId>3</depId>
         <departmentName>dental</departmentName>
      </department>
      <hospId>2</hospId>
      <hospitalName>appolo</hospitalName>
   </hospital>
   <hospital>
      <department>
         <clinic>
            <cid>82</cid>
            <clinicName>clinic82</clinicName>
            <status>0</status>
         </clinic>
         <clinic>
            <cid>92</cid>
            <clinicName>clinic92</clinicName>
            <status>0</status>
         </clinic>
         <depId>4</depId>
         <departmentName>mental</departmentName>
      </department>
      <hospId>2</hospId>
      <hospitalName>manipal</hospitalName>
   </hospital>
</hospitals>

answered Aug 2, 2012 at 7:13

Daniel Haley

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6 Comments

subbusaabu Over a year ago

xml data is not getting replaced what I do next

subbusaabu Over a year ago

Do i need to writ some code in xml file to refresh or in html any extra codes required

subbusaabu Over a year ago

one more thing if status value is '0' then we need to set status value as '1' and viceversa

subbusaabu Over a year ago

i used like this <xsl:template match="hospitals/hospital[hospId='2']/department[depId='3']/clinic[cid='9']/status"> <xsl:when test="status=0"> <status>1</status> </xsl:when> <xsl:otherwise> <status>0</status> </xsl:otherwise> </xsl:template>

subbusaabu Over a year ago

:Do I need to use any xslt engine to get the output ?

|

magix · Accepted Answer · 2012-08-02 06:49:56Z

0

It sounds as if you are trying to use XSLT to change the original XML file. XSLT cannot change the original file. You may use it to transform the XML and output it to another XML through the use of a program or script that runs XSLT.

answered Aug 2, 2012 at 6:49

magix

2244 silver badges12 bronze badges

2 Comments

subbusaabu Over a year ago

So what I need to use when I need to modify particular element in a node in xml

Andreas Over a year ago

You can a. manually replace the XML file with the XSLT output, b. use XForms (a technique for building editing Forms that is capable of changing XML files) or c. use a language that is able to trigger XSLT transformations, receive the result and save it as a file (e.g. PHP, see this answer: stackoverflow.com/questions/4946833/… )

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