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XSLT来获取需要的行，并同时改变节点值

2012-07-30 22 views
2

我目前的XML是以下几点：XSLT来获取需要的行，并同时改变节点值

<?xml version="1.0" encoding="utf-8"?> 
<Rowsets DateCreated="2012-07-17T11:57:07" EndDate="2012-07-17T11:57:07" StartDate="2012-07-17T10:57:07" Version="12.0.12 Build(9)"> 
    <Rowset> 
     <Row> 
      <CategoryID>1</CategoryID> 
      <Name>Philip</Name> 
      <City>London</City> 
      <Phone>123</Phone> 
     </Row> 
     <Row> 
      <CategoryID>2</CategoryID> 
      <Name>Derek</Name> 
      <City>Seattle</City> 
      <Phone>500</Phone> 
     </Row> 
     <Row> 
      <CategoryID>3</CategoryID> 
      <Name>Bruke</Name> 
      <City>LosAngeles</City> 
      <Phone>600</Phone> 
     </Row> 
     <Row> 
      <CategoryID>2</CategoryID> 
      <Name>Yang</Name> 
      <City>SFO</City> 
      <Phone>1233</Phone> 
     </Row> 
     <Row> 
      <CategoryID>3</CategoryID> 
      <Name>Cristina</Name> 
      <City>SanJose</City> 
      <Phone>890</Phone> 
     </Row> 
     <Row> 
      <CategoryID>4</CategoryID> 
      <Name>Meredith</Name> 
      <City>Sunnyvale</City> 
      <Phone>788</Phone> 
     </Row> 
     <Row> 
      <CategoryID>4</CategoryID> 
      <Name>Grey</Name> 
      <City>MountainView</City> 
      <Phone>456</Phone> 
     </Row> 
     <Row> 
      <CategoryID>5</CategoryID> 
      <Name>Torrence</Name> 
      <City>SAntaClara</City> 
      <Phone>432</Phone> 
     </Row> 
    </Rowset> 
</Rowsets>

现在我想只有那些<Row>其<CategoryID>为2。所以，我的XSLT是以下几点：

<?xml version="1.0" encoding="utf-8"?> 
<xsl:stylesheet xmlns:java="http://xml.apache.org/xslt/java" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" exclude-result-prefixes="java" version="1.0"> 
    <xsl:output media-type="text/xml" method="xml"/> 
    <!-- Filters refdoc based on condition and data --> 
    <xsl:template match="/"> 
     <Rowsets> 
      <xsl:for-each select="/Rowsets/Rowset"> 
       <Rowset> 
        <xsl:copy-of select="Columns"/> 
        <xsl:for-each select="Row[CategoryID = '2']"> 
         <xsl:copy-of select="."/> 
        </xsl:for-each> 
       </Rowset> 
      </xsl:for-each> 
     </Rowsets> 
    </xsl:template> 
</xsl:stylesheet>

这XSLT给了我下面的XML：

<?xml version="1.0" encoding="utf-8"?> 
<Rowsets> 
    <Rowset> 
     <Row> 
      <CategoryID>2</CategoryID> 
      <Name>Derek</Name> 
      <City>Seattle</City> 
      <Phone>500</Phone> 
     </Row> 
     <Row> 
      <CategoryID>2</CategoryID> 
      <Name>Yang</Name> 
      <City>SFO</City> 
      <Phone>1233</Phone> 
     </Row> 
    </Rowset> 
</Rowsets>

但现在我的要求就是我n个以上的新转换的XML，我也想改变类别ID为1

所以我最终的XML应该是这样的：

<?xml version="1.0" encoding="utf-8"?> 
<Rowsets> 
    <Rowset> 
     <Row> 
      <CategoryID>1</CategoryID> 
      <Name>Derek</Name> 
      <City>Seattle</City> 
      <Phone>500</Phone> 
     </Row> 
     <Row> 
      <CategoryID>1</CategoryID> 
      <Name>Yang</Name> 
      <City>SFO</City> 
      <Phone>1233</Phone> 
     </Row> 
    </Rowset> 
</Rowsets>

我不知道这怎么能以最佳的方式实现使用相同的XSLT。任何人都可以帮助我吗？

来源

2012-07-30 Soham Shah

回答

0

我不确定这是否是最好的方法，但这应该起作用。

<?xml version="1.0" encoding="utf-8"?> 
<xsl:stylesheet xmlns:java="http://xml.apache.org/xslt/java" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" exclude-result-prefixes="java" version="1.0"> 
    <xsl:output media-type="text/xml" method="xml"/> 
    <!-- Filters refdoc based on condition and data --> 
    <xsl:template match="/"> 
    <Rowsets> 
     <xsl:for-each select="/Rowsets/Rowset"> 
     <Rowset> 
      <xsl:copy-of select="Columns"/> 
      <xsl:for-each select="Row[CategoryID = '2']"> 
      <Row> 
       <xsl:apply-templates select="*" mode="Row" /> 
      </Row> 
      </xsl:for-each> 
     </Rowset> 
     </xsl:for-each> 
    </Rowsets> 
    </xsl:template> 

    <xsl:template match="*" mode="Row"> 
    <xsl:choose> 
     <xsl:when test="current()[name()='CategoryID']"> 
     <xsl:copy> 
      <xsl:text>1</xsl:text> 
     </xsl:copy> 
     </xsl:when> 
     <xsl:otherwise> 
     <xsl:copy-of select="." /> 
     </xsl:otherwise> 
    </xsl:choose> 
    </xsl:template> 
</xsl:stylesheet>

输出将是

<?xml version="1.0" encoding="UTF-8"?> 
<Rowsets> 
    <Rowset> 
    <Row> 
     <CategoryID>1</CategoryID> 
     <Name>Derek</Name> 
     <City>Seattle</City> 
     <Phone>500</Phone> 
    </Row> 
    <Row> 
     <CategoryID>1</CategoryID> 
     <Name>Yang</Name> 
     <City>SFO</City> 
     <Phone>1233</Phone> 
    </Row> 
    </Rowset> 
</Rowsets>

来源

2012-07-31 00:00:07 james31rock

2

该转化：

<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="Row[not(CategoryID = 2)]"/> 

<xsl:template match="CategoryID/text()">1</xsl:template> 
</xsl:stylesheet>

当所提供的XML文档应用：

<Rowsets DateCreated="2012-07-17T11:57:07" EndDate="2012-07-17T11:57:07" StartDate="2012-07-17T10:57:07" Version="12.0.12 Build(9)"> 
    <Rowset> 
     <Row> 
      <CategoryID>1</CategoryID> 
      <Name>Philip</Name> 
      <City>London</City> 
      <Phone>123</Phone> 
     </Row> 
     <Row> 
      <CategoryID>2</CategoryID> 
      <Name>Derek</Name> 
      <City>Seattle</City> 
      <Phone>500</Phone> 
     </Row> 
     <Row> 
      <CategoryID>3</CategoryID> 
      <Name>Bruke</Name> 
      <City>LosAngeles</City> 
      <Phone>600</Phone> 
     </Row> 
     <Row> 
      <CategoryID>2</CategoryID> 
      <Name>Yang</Name> 
      <City>SFO</City> 
      <Phone>1233</Phone> 
     </Row> 
     <Row> 
      <CategoryID>3</CategoryID> 
      <Name>Cristina</Name> 
      <City>SanJose</City> 
      <Phone>890</Phone> 
     </Row> 
     <Row> 
      <CategoryID>4</CategoryID> 
      <Name>Meredith</Name> 
      <City>Sunnyvale</City> 
      <Phone>788</Phone> 
     </Row> 
     <Row> 
      <CategoryID>4</CategoryID> 
      <Name>Grey</Name> 
      <City>MountainView</City> 
      <Phone>456</Phone> 
     </Row> 
     <Row> 
      <CategoryID>5</CategoryID> 
      <Name>Torrence</Name> 
      <City>SAntaClara</City> 
      <Phone>432</Phone> 
     </Row> 
    </Rowset> 
</Rowsets>

产生想要的，正确的结果：

<Rowsets DateCreated="2012-07-17T11:57:07" EndDate="2012-07-17T11:57:07" StartDate="2012-07-17T10:57:07" Version="12.0.12 Build(9)"> 
    <Rowset> 
     <Row> 
     <CategoryID>1</CategoryID> 
     <Name>Derek</Name> 
     <City>Seattle</City> 
     <Phone>500</Phone> 
     </Row> 
     <Row> 
     <CategoryID>1</CategoryID> 
     <Name>Yang</Name> 
     <City>SFO</City> 
     <Phone>1233</Phone> 
     </Row> 
    </Rowset> 
</Rowsets>

说明：的identity rule

  1. 正确使用和覆盖。

  2. 正确使用模板和模式匹配。

来源

2012-07-31 01:24:48

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