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请建议获取通过xslt集合收集的每个文档的绝对路径。如何从集合中获取文件的绝对文档路径
发布脚本能够得到所需绝对路径,但我已经用了两个集合(它可能冒不必要的内存来存储所有文章的信息两次,一次采集,收集信息等一个收集document-uri()
S)。
个XML:
d:/DocumentPath/Project-01/2016/ABC/Test.xml
<article>
<title>First article</title>
<tag1>The tag 1</tag1>
<tag3>The tag 3</tag3>
</article>
d:/ DocumentPath /项目-01/2016/DEF /测试。 XML
<article>
<title>Second article</title>
<tag2>The tag 2</tag2>
<tag3>The tag 3</tag3>
</article>
等个XML ....
XSLT 2.0:
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" omit-xml-declaration="yes" indent="yes"/>
<xsl:variable name="varDocuments">
<xsl:copy-of select="collection('file:///D:/DocumentPath/Project-01/2016/?select=*.xml;recurse=yes')
[matches(document-uri(.), '2016/([A-z]+)/.*?.xml')]"/>
</xsl:variable>
<xsl:variable name="varDocuments1">
<xsl:copy-of select="collection('file:///D:/DocumentPath/Project-01/2016/?select=*.xml;recurse=yes')
[matches(document-uri(.), '2016/([A-z]+)/.*?.xml')]/document-uri(.)"/>
</xsl:variable>
<xsl:template match="@*|node()">
<xsl:copy><xsl:apply-templates select="@*|node()"/></xsl:copy>
</xsl:template>
<xsl:template match="/">
<Table border="1">
<TR><TH>Position</TH><TH>Title</TH><TH>Tag1</TH><TH>Tag2</TH><TH>Tag3</TH><TH>Tag4</TH><TH>Path</TH></TR>
<xsl:for-each select="$varDocuments">
<xsl:for-each select="article">
<TR>
<xsl:variable name="varPos" select="position()"/>
<td><xsl:value-of select="position()"/></td>
<td><xsl:value-of select="title"/></td>
<td><xsl:value-of select="count(descendant::tag1)"/></td>
<td><xsl:value-of select="count(descendant::tag2)"/></td>
<td><xsl:value-of select="count(descendant::tag3)"/></td>
<td><xsl:value-of select="count(descendant::tag4)"/></td>
<td><xsl:value-of select="normalize-space(tokenize($varDocuments1, 'file:/')[position()=$varPos + 1])"/></td>
</TR>
</xsl:for-each>
</xsl:for-each>
</Table>
</xsl:template>
</xsl:stylesheet>
要求的结果:
<Table border="1">
<TR>
<TH>Position</TH>
<TH>Title</TH>
<TH>Tag1</TH>
<TH>Tag2</TH>
<TH>Tag3</TH>
<TH>Tag4</TH>
<TH>Path</TH>
</TR>
<TR>
<td>1</td>
<td>First article</td>
<td>1</td>
<td>0</td>
<td>1</td>
<td>0</td>
<td>D:/DocumentPath/Project-01/2016/ABC/Test.xml</td>
</TR>
<TR>
<td>2</td>
<td>Second article</td>
<td>0</td>
<td>1</td>
<td>1</td>
<td>0</td>
<td>D:/DocumentPath/Project-01/2016/DEF/Test.xml</td>
</TR>
<TR>
<td>3</td>
<td>Third article</td>
<td>1</td>
<td>0</td>
<td>0</td>
<td>2</td>
<td>D:/DocumentPath/Project-01/2016/GHI/Test.xml</td>
</TR>
</Table>
谢谢你的完美建议,先生,得到所需的绝对路径,加上一个。 –