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我有问题得到一个非常简单的XSL:排序工作。我确信解决方案是微不足道的,但从这里和其他资源类似的问题的例子没有得到我的解决方案。XSLT为每个与排序不按预期工作
示例XML
<Package version="1.0">
<Payload>
<Filenames>
<Filename>MDP_939529_0_20151006104742_2.tif</Filename>
<Filename>MDP_939529_0_20151006104742_3.tif</Filename>
<Filename>MDP_939529_0_20151006104742_1.tif</Filename>
</Filenames>
<DocumentType>FOO</DocumentType>
<Timestamp>2015-10-26T15:42:04.902-07:00</Timestamp>
<PayloadContext>
<Item name="Company ID">0</Item>
<Item name="Office ID">0</Item>
</PayloadContext>
</Payload>
</Package>
我的XSLT是
<?xml version='1.0' ?>
<xsl:stylesheet
version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output
method="xml"
version="1.0"
encoding="UTF-8"
indent="yes"
omit-xml-declaration="yes"/>
<xsl:strip-space elements="*"/>
<!-- global template to copy everything that doesn't match the other templates -->
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/MDXPackage/Payload/Filenames">
<Filenames>
<xsl:for-each select="//MDXPackage/Payload/Filenames/Filename">
<xsl:sort select="Filename" data-type="text" />
<Filename>
<xsl:value-of select="//MDXPackage/Payload/Filenames/Filename"/>
</Filename>
</xsl:for-each>
</Filenames>
</xsl:template>
</xsl:stylesheet>
我的结果中包含
<Filenames>
<Filename>MDP_939529_0_20151006104742_2.tif</Filename>
<Filename>MDP_939529_0_20151006104742_2.tif</Filename>
<Filename>MDP_939529_0_20151006104742_2.tif</Filename>
</Filenames>
,而不是期望
<Filenames>
<Filename>MDP_939529_0_20151006104742_1.tif</Filename>
<Filename>MDP_939529_0_20151006104742_2.tif</Filename>
<Filename>MDP_939529_0_20151006104742_3.tif</Filename>
</Filenames>
在此先感谢
大卫
谢谢。完美的作品 – DLoysen
@DLoysen,不客气 –