[java] нужен XML Parser/XPath

data match {
  case <string>{content @ _*}</string> => content
}

<xsl:template match="string">
  <xsl:value-of select="." />
</xsl:template>

Welcome to Scala version 2.8.1.final (Java HotSpot(TM) Client VM, Java 1.6.0_22).
Type in expressions to have them evaluated.
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scala> def extract(x: xml.Node): xml.NodeSeq = x match {
     |   case <string>{content @ _*}</string> => content
     | }
extract: (x: scala.xml.Node)scala.xml.NodeSeq

scala> val x = <string>string1<string>string2</string>string1</string>
x: scala.xml.Elem = <string>string1<string>string2</string>string1</string>

scala> extract(x)
res0: scala.xml.NodeSeq = NodeSeq(string1, <string>string2</string>, string1)

<?xml version="1.0" encoding="utf-8"?>
<resources>
    <string name="app_name">SPB TV</string>
    <string name="yes">yes</string>
</resources>

<?xml version="1.0" encoding="utf-8"?>
<resources>
    <string>
        <src>SPB TV</src>
        <dst>СПБ ТВ</dst>
    </string>    
</resources>

<?xml version="1.0" encoding="utf-8"?>
<xsl:transform version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="xml" version="1.0" encoding="utf-8" indent="yes"/>
    <xsl:variable name="with" select="'translation.xml'"/>

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

    <xsl:template match="/resources/string">
        <xsl:copy>
            <xsl:apply-templates select="@*"/>
            <xsl:variable name="info"
                          select="document($with)/resources/string[src/node()=current()/node()]/."/>
            <xsl:choose>
                <xsl:when test="$info">
                    <xsl:copy-of select="$info/dst/node()"/>
                </xsl:when>
                <xsl:otherwise>
                    <xsl:copy-of select="current()/node()"/>
                </xsl:otherwise>
            </xsl:choose>
        </xsl:copy>
    </xsl:template>

</xsl:transform>