我想编写一个xslt来将一个xml文件转换为另一个xml文件。源XML文件类似于以下
<orgs>
<organization revenue="10000">
<name>foo</name>
</organization>
<organization parent="foo">
<name>foo2</name>
</organization>
<organization parent="foo2">
<name>foo3</name>
</organization>
</orgs>
输出xml应如下
<orgo>
<organization revenue="10000">
<name>foo</name>
<organization>
<name>foo2</name>
<organization><name>foo3</name></organization>
</organization>
</organization>
</orgo>
到目前为止,我已尝试按如下方式编写xsl
答案 0 :(得分:1)
此转化:
<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()[1]"/>
<xsl:apply-templates select="following-sibling::node()[1]"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
应用于提供的XML文档:
<orgs>
<organization revenue="10000">
<name>foo</name>
</organization>
<organization parent="foo">
<name>foo2</name>
</organization>
<organization parent="foo2">
<name>foo3</name>
</organization>
</orgs>
生成所需的正确输出:
<orgs>
<organization>
<name>foo</name>
<organization>
<name>foo2</name>
<organization>
<name>foo3</name>
</organization>
</organization>
</organization>
</orgs>
如果<organization>
元素的顺序是随机的,请参阅以下XML文档:
<orgs>
<organization parent="foo2">
<name>foo3</name>
</organization>
<organization parent="foo">
<name>foo2</name>
</organization>
<organization revenue="10000">
<name>foo</name>
</organization>
</orgs>
此转换产生想要的结果:
<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="/*">
<xsl:copy>
<xsl:apply-templates select="organization[not(@parent)]"/>
</xsl:copy>
</xsl:template>
<xsl:template match="organization">
<xsl:copy>
<xsl:copy-of select="node()"/>
<xsl:apply-templates select="../organization[@parent=current()/name]"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>