xml - XSLT newbie, replacing values -

my mistake, make clear:

i have simple question, new xslt:

i have 2 xml files:

usa1.xml:

<?xml version="1.0" encoding="utf-8"?> <country>   <state name="ca">     <city name="sunnyvale" county="sant clara">       <street number="123">         el comino ave.       </street>     </city>     <city name="san jose" county="sant clara">       <street number="345">         de anza ave.       </street>     </city>     <city name="palo alto" county="sant clara">       <street number="789">         shoreline ave.       </street>     </city>    </state> </country> 

usa2.xml:

<?xml version="1.0" encoding="utf-8"?> <country>   <state name="ca">     <city name="sunnyvale" county="sant clara">       <street number="999">         homestead ave.       </street>     </city>     <city name="san jose" county="sant clara">       <street number="888">         airport ave.       </street>     </city>   </state> </country> 

i want use xslt replace values , attributes of city sunnyvale , san jose in usa1.xml data usa2.xml.

idea output usa4.xml:

usa4.xml:

<?xml version="1.0" encoding="utf-8"?> <country>   <state name="ca">     <city name="sunnyvale" county="sant clara">       <street number="999">         homestead ave.       </street>     </city>     <city name="san jose" county="sant clara">       <street number="888">         airport ave.       </street>     </city>     <city name="palo alto" county="sant clara">       <street number="789">         shoreline ave.       </street>     </city>    </state> </country> 

how can this?

i tried following xslt, returned output not expecting:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/xsl/transform">   <xsl:output method="xml" indent="yes"/>   <xsl:param name="usaxml" select="'usa1.xml'" />   <xsl:variable name="address" select="document($usaxml)//" />    <xsl:template match="/">     <xsl:attribute name="number">        <xsl:value-of select="$address/@street" />     </xsl:attribute>   </xsl:template>    <xsl:template match="@*|node()">     <xsl:copy>       <xsl:apply-templates select="@*|node()"/>     </xsl:copy>   </xsl:template> </xsl:stylesheet> 

i use key cross reference between 2 files admittedly code becomes bit convoluted xslt 1.0 if need use keys between documents:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/xsl/transform">  <xsl:output method="xml" indent="yes"/>  <xsl:key name="k1" match="state/city" use="concat(../@name, '|', @name)"/>  <xsl:param name="usaxml" select="'test2013050302.xml'" /> <xsl:variable name="address" select="document($usaxml)" />  <xsl:template match="@* | node()">   <xsl:copy>     <xsl:apply-templates select="@* | node()"/>   </xsl:copy> </xsl:template>  <xsl:template match="state/city">   <xsl:copy>     <xsl:variable name="this" select="."/>     <xsl:for-each select="$address">       <xsl:variable name="ref-data" select="key('k1', concat($this/../@name, '|', $this/@name))"/>       <xsl:choose>         <xsl:when test="$ref-data">           <xsl:apply-templates select="$ref-data/@* | $ref-data/node()"/>         </xsl:when>         <xsl:otherwise>           <xsl:apply-templates select="$this/@* | $this/node()"/>         </xsl:otherwise>       </xsl:choose>     </xsl:for-each>   </xsl:copy> </xsl:template>  </xsl:stylesheet> 

so primary input being

<?xml version="1.0" encoding="utf-8"?> <country>   <state name="ca">     <city name="sunnyvale" county="sant clara">       <street number="123">         el comino ave.       </street>     </city>     <city name="san jose" county="sant clara">       <street number="345">         de anza ave.       </street>     </city>     <city name="palo alto" county="sant clara">       <street number="789">         shoreline ave.       </street>     </city>    </state> </country> 

and parameter file test2013050302.xml being

<?xml version="1.0" encoding="utf-8"?> <country>   <state name="ca">     <city name="sunnyvale" county="sant clara">       <street number="999">         homestead ave.       </street>     </city>     <city name="san jose" county="sant clara">       <street number="888">         airport ave.       </street>     </city>   </state> </country> 

the transformation result is

<country>   <state name="ca">     <city name="sunnyvale" county="sant clara">       <street number="999">         homestead ave.       </street>     </city>     <city name="san jose" county="sant clara">       <street number="888">         airport ave.       </street>     </city>     <city name="palo alto" county="sant clara">       <street number="789">         shoreline ave.       </street>     </city>    </state> </country> 

using xslt 2.0 processor , xslt 2.0 vastly simplify template

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/xsl/transform">  <xsl:output method="xml" indent="yes"/>  <xsl:key name="k1" match="state/city" use="concat(../@name, '|', @name)"/>  <xsl:param name="usaxml" select="'test2013050302.xml'" /> <xsl:variable name="address" select="document($usaxml)" />  <xsl:template match="@* | node()">   <xsl:copy>     <xsl:apply-templates select="@* | node()"/>   </xsl:copy> </xsl:template>  <xsl:template match="state/city">   <xsl:copy>     <xsl:variable name="ref-data" select="key('k1', concat(../@name, '|', @name), $address)"/>     <xsl:choose>       <xsl:when test="$ref-data">         <xsl:apply-templates select="$ref-data/(@*, node())"/>       </xsl:when>       <xsl:otherwise>         <xsl:apply-templates select="@* , node()"/>       </xsl:otherwise>     </xsl:choose>   </xsl:copy> </xsl:template>  </xsl:stylesheet>