使用XSLT将XML文件转换为另一个XML文件

XML 文件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文件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>

如何使用xslt将xml文件1转换为xml fle 2？
我想将地址表示为单行,并将一个新属性[country-Australia]添加到行尾.我做了其余的.我正在努力地址线

XSLT文件：

<?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>

这种转变：

<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>

当应用于提供的XML文档时：

<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>

产生想要的,正确的结果：

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

说明：使用和覆盖identity rule.

使用XSLT将XML文件转换为另一个XML文件

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