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현재 해결해야 할 문제는 xslt를 통해 차이점과 유사점이있는 두 xml 파일의 비교를 생성하는 것입니다. 예상대로 I 출력 작품 파일 XSLT에 두번째 파일을 포함하는 경우xslt에 XML 스 니펫을 매개 변수로 전달
가<?xml version="1.0" encoding="utf-8" ?>
<Stats Date="2011-01-01">
<Player Rank="1">
<GP>39</GP>
<G>32</G>
<A>33</A>
<PlusMinus>20</PlusMinus>
<PIM>29</PIM>
<PP>10</PP>
<SH>1</SH>
<GW>3</GW>
<Shots>0</Shots>
<ShotPctg>154</ShotPctg>
<TOIPerGame>20.8</TOIPerGame>
<ShiftsPerGame>21:54</ShiftsPerGame>
<FOWinPctg>22.6</FOWinPctg>
</Player>
</Stats>
번째 파일
<?xml version="1.0" encoding="utf-8" ?>
<Stats Date="2011-01-01">
<Player Rank="2">
<Name>John Smith</Name>
<Team>NY</Team>
<Pos>D</Pos>
<GP>38</GP>
<G>32</G>
<A>33</A>
<PlusMinus>15</PlusMinus>
<PIM>29</PIM>
<PP>10</PP>
<SH>1</SH>
<GW>4</GW>
<Shots>0</Shots>
<ShotPctg>158</ShotPctg>
<TOIPerGame>20.8</TOIPerGame>
<ShiftsPerGame>21:54</ShiftsPerGame>
<FOWinPctg>22.6</FOWinPctg>
</Player>
</Stats>
보이는 것처럼
예를 들어 첫 번째 XML 파일 같습니다
<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="vrtfDoc2">
<Stats Date="2011-01-01">
<Player Rank="2">
<Name>John Smith</Name>
<Team>NY</Team>
<Pos>D</Pos>
<GP>38</GP>
<G>32</G>
<A>33</A>
<PlusMinus>15</PlusMinus>
<PIM>29</PIM>
<PP>10</PP>
<SH>1</SH>
<GW>4</GW>
<Shots>0</Shots>
<ShotPctg>158</ShotPctg>
<TOIPerGame>20.8</TOIPerGame>
<ShiftsPerGame>21:54</ShiftsPerGame>
<FOWinPctg>22.6</FOWinPctg>
</Player>
</Stats>
</xsl:param>
<xsl:variable name="vDoc2" select=
"document('')/*/xsl:param[@name='vrtfDoc2']/*"/>
<xsl:template match="node()|@*" name="identity">
<xsl:param name="pDoc2"/>
<xsl:copy>
<xsl:apply-templates select="node()|@*">
<xsl:with-param name="pDoc2" select="$pDoc2"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="/">
<xsl:apply-templates select="*">
<xsl:with-param name="pDoc2" select="$vDoc2"/>
</xsl:apply-templates>
-----------------------
<xsl:apply-templates select="$vDoc2">
<xsl:with-param name="pDoc2" select="/*"/>
</xsl:apply-templates>
</xsl:template>
<xsl:template match="Player/*">
<xsl:param name="pDoc2"/>
<xsl:if test=
"not(. = $pDoc2/*/*[name()=name(current())])">
<xsl:call-template name="identity"/>
</xsl:if>
</xsl:template>
<xsl:template match="Name|Team|Pos" priority="20"/>
</xsl:stylesheet>
다음 C# 코드를 사용하는 경우 :
나는 내 인생 I, 나는이 변환에서 빈 출력을 얻을private string Transform(string xml, string xml2, string xsl) {
StringWriter writer = new StringWriter();
XslCompiledTransform t = new XslCompiledTransform(true);
XsltSettings settings = new XsltSettings(true, false);
XmlTextReader xmlReader = new XmlTextReader(xml);
XmlDocument doc1 = new XmlDocument();
// populate as needed e.g.
doc1.Load(xml2);
XmlTextReader xslReader = new XmlTextReader(xsl);
t.Load(xslReader, settings, null);
//Pass parameter value to xslt from code
XsltArgumentList argumentList = new XsltArgumentList();
argumentList.AddParam("vrtfDoc2", "", doc1);
t.Transform(xmlReader, argumentList, writer);
return writer.ToString();
}
에 XSLT
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:param name="vrtfDoc2" />
<xsl:variable name="vDoc2" select=
"document('')/*/xsl:param[@name='vrtfDoc2']/*"/>
<xsl:template match="node()|@*" name="identity">
<xsl:param name="pDoc2"/>
<xsl:copy>
<xsl:apply-templates select="node()|@*">
<xsl:with-param name="pDoc2" select="$pDoc2"/>
</xsl:apply-templates>
</xsl:copy>
</xsl:template>
<xsl:template match="/">
<xsl:apply-templates select="*">
<xsl:with-param name="pDoc2" select="$vDoc2"/>
</xsl:apply-templates>
-----------------------
<xsl:apply-templates select="$vDoc2">
<xsl:with-param name="pDoc2" select="/*"/>
</xsl:apply-templates>
</xsl:template>
<xsl:template match="Player/*">
<xsl:param name="pDoc2"/>
<xsl:if test=
"not(. = $pDoc2/*/*[name()=name(current())])">
<xsl:call-template name="identity"/>
</xsl:if>
</xsl:template>
<xsl:template match="Name|Team|Pos" priority="20"/>
</xsl:stylesheet>
변화는 C# 방법에서 포함 된 XML을 제거
private string Transform(string xml, string xml2, string xsl) {
StringWriter writer = new StringWriter();
XslCompiledTransform t = new XslCompiledTransform(true);
XsltSettings settings = new XsltSettings(true, false);
XmlTextReader xmlReader = new XmlTextReader(xml);
XmlTextReader xslReader = new XmlTextReader(xsl);
t.Load(xslReader, settings, null);
t.Transform(xmlReader, null, writer);
return writer.ToString();
}
이유를 이해할 수 없습니다. 내가 디버거를 사용하여 두 버전을 통해 강화하지했습니다 및 매개 변수 값이 모두 occassions에 동일 보이지만 매개 변수 전달 버전은 XSLT에 다음 코드를 칠 때 어떤 변화가 발생합니다
<xsl:apply-templates select="$vDoc2">
<xsl:with-param name="pDoc2" select="/*"/>
</xsl:apply-templates>
</xsl:template>
<xsl:template match="Player/*">
<xsl:param name="pDoc2"/>
<xsl:if test=
"not(. = $pDoc2/*/*[name()=name(current())])">
<xsl:call-template name="identity"/>
</xsl:if>
</xsl:template>
어떤 도움이나 제안이 많은 것 appeciated.