Question

我可以在Java中使用XSLT 1.0，如下面的示例所示： -

copy.xml

<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="identityxfm.xsl"?>
<catalog>
    <book id="bk101">
        <author>Gambardella, Matthew</author>
        <title>XML Developer's Guide</title>
        <genre>Computer</genre>
        <price>44.95</price>
        <publish_date>2000-10-01</publish_date>        
        <description>An in-depth look at creating applications with
 XML.</description>
    </book>
    <book id="bk102">
        <author>Ralls, Kim</author>
        <title>Midnight Rain</title>
        <genre>Fantasy</genre>
        <price>5.95</price>
        <publish_date>2000-12-16</publish_date>
        <description>A former architect battles corporate zombies,
 an evil sorceress, and her own childhood to become queen of the
 world.</description>
    </book>
    <book id="bk103">
        <author>Corets, Eva</author>
        <title>Maeve Ascendant</title>
        <genre>Fantasy</genre>
        <price>5.95</price>
        <publish_date>2000-11-17</publish_date>
        <description>After the collapse of a nanotechnology society
 in England, the young survivors lay the foundation for a new 
society.</description>
    </book>
</catalog>

copy.xsl

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
      xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >

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

</xsl:stylesheet>

Copy.java

package com.data.transform;

import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.stream.StreamResult;
import javax.xml.transform.stream.StreamSource;

public class Copy {

    public static void main(String args[]) throws Exception {
        StreamSource source = new StreamSource("copy.xml");
        StreamSource stylesource = new StreamSource("copy.xsl");

        TransformerFactory factory = TransformerFactory.newInstance();
        Transformer transformer = factory.newTransformer(stylesource);

        StreamResult result = new StreamResult(System.out);
        transformer.transform(source, result);
      }
}

输出

<?xml version="1.0" encoding="UTF-8"?><?xml-stylesheet type="text/xsl" href="identityxfm.xsl"?><catalog>
    <book id="bk101">
        <author>Gambardella, Matthew</author>
        <title>XML Developer's Guide</title>
        <genre>Computer</genre>
        <price>44.95</price>
        <publish_date>2000-10-01</publish_date>        
        <description>An in-depth look at creating applications with
 XML.</description>
    </book>
    <book id="bk102">
        <author>Ralls, Kim</author>
        <title>Midnight Rain</title>
        <genre>Fantasy</genre>
        <price>5.95</price>
        <publish_date>2000-12-16</publish_date>
        <description>A former architect battles corporate zombies,
 an evil sorceress, and her own childhood to become queen of the
 world.</description>
    </book>
    <book id="bk103">
        <author>Corets, Eva</author>
        <title>Maeve Ascendant</title>
        <genre>Fantasy</genre>
        <price>5.95</price>
        <publish_date>2000-11-17</publish_date>
        <description>After the collapse of a nanotechnology society
 in England, the young survivors lay the foundation for a new 
society.</description>
    </book>
</catalog>

但是，现在我想在Java中使用XSLT 2.0和XSLT 3.0（如xsl:analyze-string，xsl:try等）中包含的一些东西。我怎么能这样做？

Answer 1

从Maven或Sourceforge获取Saxon 9 HE并将其放在类路径上，然后在9.8或XSLT 3.0支持之前对Saxon 9.x提供XSLT 2.0支持（流，高阶函数除外，{{1使用9.8，模式感知，向后兼容性）。要获得完整的XSLT 3.0支持，您需要to download Saxon 9 PE or EE from Saxonica并将其与您购买的license或您在类路径上请求的试用许可证放在一起。

如何在Java中使用XSLT 2.0和XSLT 3.0？

1 个答案: