Converting XML file to another XML file using XSLT

Question

XML file 1:

<?xml version="1.0"?>
<rentalProperties>
    <property contact ="1">
        <type>House </type>
        <price>420</price>
        <address>
            <streetNo>1</streetNo>
            <street>Wavell Street</street>
            <suburb>Box Hill</suburb>
            <state>VIC</state>
            <zipcode>3128</zipcode> 
        </address>
        <numberOfBedrooms>3</numberOfBedrooms>
        <numberOfBathrooms>1</numberOfBathrooms> 
        <garage>1</garage>   
    </property>

XML file 2:

<?xml version="1.0"?>
<rentalProperties>
    <property contact ="1">
        <type>House </type>
        <price>420</price>
        <address>1 wavell street,Box Hill,VIC,Australia</address>
        <numberOfBedrooms>3</numberOfBedrooms>
        <numberOfBathrooms>1</numberOfBathrooms> 
        <garage>1</garage>     
    </property>

How should i convert xml file 1 to xml fle 2 using xslt? i want to represent the address as the single line and add a new attribute [country- Australia] to end of the line. i did the rest of it . i'm struggling with address line

XSLT file:

<?xml version="1.0" encoding="iso-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" type="text/css" href="style.css">
    <xsl:template match="/">
        <rentalProperties>
            <property>
                <xsl:attribute name="contact"><xsl:value-of select='@contact'/></xsl:attribute>    
                <type><xsl:value-of select="type"/></type>
                <price><xsl:value-of select="price"/></price>
                <numberOfBedrooms><xsl:value-of select="numberOfBedrooms"/></numberOfBedrooms>
                <numberOfBathrooms><xsl:value-of select="numberOfBathrooms"/></numberOfBathrooms>
                <garage><xsl:value-of select="garage"/></garage>    
            </property>    
        </rentalProperties>    
    </xsl:template>
</xsl:stylesheet>

Dimitre Novatchev · Accepted Answer · 2011-05-14 18:18:54Z

32

This 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="address">
  <xsl:copy>
   <xsl:value-of select=
   "concat(streetNo, ' ', street, ',',
           suburb,',', state,', Australia')
   "/>
  </xsl:copy>
 </xsl:template>
 <xsl:template match="address/node()"/>
</xsl:stylesheet>

when applied on the provided XML document:

<rentalProperties>
    <property contact ="1">
        <type>House </type>
        <price>420</price>
        <address>
            <streetNo>1</streetNo>
            <street>Wavell Street</street>
            <suburb>Box Hill</suburb>
            <state>VIC</state>
            <zipcode>3128</zipcode>
        </address>
        <numberOfBedrooms>3</numberOfBedrooms>
        <numberOfBathrooms>1</numberOfBathrooms>
        <garage>1</garage>
    </property>
</rentalProperties>

produces the wanted, correct result:

<rentalProperties>
   <property contact="1">
      <type>House </type>
      <price>420</price>
      <address>1 Wavell Street,Box Hill,VIC, Australia</address>
      <numberOfBedrooms>3</numberOfBedrooms>
      <numberOfBathrooms>1</numberOfBathrooms>
      <garage>1</garage>
   </property>
</rentalProperties>

Explanation: Using and overriding the identity rule.

answered May 14, 2011 at 18:18

Dimitre Novatchev

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2 Comments

mgouin Over a year ago

Is it possible to also include the xml header in the output: <?xml version="1.0" encoding="iso-8859-1"?>

Dimitre Novatchev Over a year ago

@mgouin, It is called "XML declaration". Yes, just remove the omit-xml-declaration="yes" attribute from the <xsl:output> instruction

Osiris76 · Accepted Answer · 2011-05-14 15:35:28Z

4

You could introduce a new template for the address block using

<xsl:template match="address">
    <xsl:value-of select="streetNo" />
    <xsl:text> </xsl:text>
    <xsl:value-of select="street" />
    <xsl:text>,</xsl:text>
    <xsl:value-of select="suburb" />
    <xsl:text>,</xsl:text>
    <xsl:value-of select="state" />
    <xsl:text>,</xsl:text>
    <xsl:value-of select="zipcode" />
</xsl:template>

and call it with

<xsl:apply-templates select="address" />

before the <numberOfBedrooms> element. This can also be done using the concat function, whereas the correct syntax I don't remember right now.

answered May 14, 2011 at 15:35

Osiris76

1,2549 silver badges5 bronze badges

Comments

actionshrimp · Accepted Answer · 2011-05-14 15:30:28Z

1

You can try something like:

<address>
    <xsl:for-each select="address/*">
       <xsl:value-of select="."/>, 
    </xsl:for-each>
    Australia
</address>

This loops over all the children of the address tag in xml1.

answered May 14, 2011 at 15:30

actionshrimp

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Comments

raj · Accepted Answer · 2016-01-28 16:09:46Z

-3

<rentalProperties>
    <property contact="1">
        <type>House </type>
        <price>420</price>
        <address>1 Wavell Street,Box Hill,VIC,3128</address>
        <numberOfBedrooms>3</numberOfBedrooms>
        <numberOfBathrooms>1</numberOfBathrooms> 
        <garage>1</garage>   
    </property>
</rentalProperties>

answered Jan 28, 2016 at 16:09

raj

1

2 Comments

Benjamin W. Over a year ago

Code-only answers can almost always be improved by adding a few explanatory sentences; you can edit your answer to do so.

Marco Roy Over a year ago

What's the solution here?

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