First, let me provide the sample xml so you guys will be clear what am after.
<a>1</a>
<b>1</b>
<c>1</c>
<d>1</d>
<e>1</e>
<f>1</f>
Is is possible to copy node from a to b and from e to f. I need to neglect node c and d.
There is <xsl:copy>
which can copy the elements, but I need to get particular element out of original XML.
Thank you.
答案 0 :(得分:0)
确定您可以删除所需的元素。只需在身份转换后在指定元素上写空模板。
源XML
<root>
<a>1</a>
<b>1</b>
<c>1</c>
<d>1</d>
<e>1</e>
<f>1</f>
</root>
XSLT 1.0
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:strip-space elements="*"/>
<!-- Identity Transform -->
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<!-- Empty Template to Remove Elements -->
<xsl:template match="c|d"/>
</xsl:stylesheet>
输出XML
<root>
<a>1</a>
<b>1</b>
<e>1</e>
<f>1</f>
</root>
或者,选择要保留的特定节点:
XSLT 1.0
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:strip-space elements="*"/>
<!-- Root Template Match -->
<xsl:template match="root">
<xsl:copy>
<xsl:apply-templates select="a|b|e|f"/>
</xsl:copy>
</xsl:template>
<!-- Select Particular Elements -->
<xsl:template match="a|b|e|f">
<xsl:copy-of select="."/>
</xsl:template>
</xsl:stylesheet>