Question

我正在开发一个XSLT转换，它只能在XPath 1.0中使用xslt 1.0。假设我有以下xml，我想选择original_file_name属性以“.prt”结尾的对象。我怎么能这样做？

<object>
    <object>
        <attribute name="original_file_name">file1.qaf</attribute>
        <attribute name="otherAttribute1">val</attribute>
        <attribute name="otherAttribute2">val</attribute>
    </object>
    <object>
        <attribute name="original_file_name">file2.tso</attribute>
        <attribute name="otherAttribute1">Second guess</attribute>
        <attribute name="otherAttribute2">val</attribute>
    </object>
    <object>
        <attribute name="original_file_name">file3.prt</attribute>
        <attribute name="otherAttribute1">First guess!</attribute>
        <attribute name="otherAttribute2">val</attribute>
    </object>
</object>

对于XSLT 2.0，我找到了一个解决方案，但我无法摆脱XPath 1.0中不支持的索引。你有任何解决方案的提示吗？

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:fn="http://www.w3.org/2005/xpath-functions">
    <xsl:output method="xml" encoding="iso-8859-1" indent="yes"/>
    <xsl:template match="/object">
        <xsl:variable name="fileIndex">
            <xsl:call-template name="identifyFile">
                <xsl:with-param name="fileNames" select="object/attribute[@name='original_file_name']"/>
            </xsl:call-template>
        </xsl:variable>
        <xsl:variable name="fileObj" select="object[position()=$fileIndex]"/>

        <!-- do sth. with fileObj -->
        <xsl:copy-of select="$fileObj" />
    </xsl:template>
    <xsl:template name="identifyFile">
        <xsl:param name="fileNames"/>
        <xsl:choose>
            <!-- add other file extensions here -->
            <xsl:when test="$fileNames['.prt'=substring(., string-length() - 3)]">
                <xsl:value-of select="index-of($fileNames, $fileNames['.prt'=substring(., string-length() - 3)])" />
            </xsl:when>
            <xsl:when test="$fileNames['.tso'=substring(., string-length() - 3)]">
                <xsl:value-of select="index-of($fileNames, $fileNames['.tso'=substring(., string-length() - 3)])" />
            </xsl:when>
            <!-- if no known file extension has been found, just take the default one. -->
            <xsl:otherwise>
                <xsl:value-of select="1"/>
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>
</xsl:stylesheet>

编辑：在第一步中过多地简化了问题:)。我添加了第二个文件扩展名，如果找不到第一个文件扩展名，应该搜索它。因此，如果在示例中没有'.prt'文件，则应该采用'.tso'文件。

Answer 1

如果您只想根据条件选择XPath表达式

object[attribute[@name='original_file_name' and substring(., string-length() - 3) = '.prt']]

copy-of中的

就足够了。使用您编辑的样本我认为您只需要

   <xsl:template match="/object">
     <xsl:variable name="fileNames" select="object/attribute[@name='original_file_name']"/>
        <xsl:choose>
            <!-- add other file extensions here -->
            <xsl:when test="$fileNames['.prt'=substring(., string-length() - 3)]">
                <xsl:copy-of select="$fileNames['.prt'=substring(., string-length() - 3)]" />
            </xsl:when>
            <xsl:when test="$fileNames['.tso'=substring(., string-length() - 3)]">
                <xsl:copy-of select="$fileNames['.tso'=substring(., string-length() - 3)]" />
            </xsl:when>
            <!-- if no known file extension has been found, just take the default one. -->
            <xsl:otherwise>
                <xsl:copy-of select="$fileNames[1]"/>
            </xsl:otherwise>
        </xsl:choose>
    </xsl:template>

XPath / XSLT 1.0：根据子元素值标识元素

1 个答案: