输入:
输入XML
<Root>
<Number>1</Number>
<Reference>100</Reference>
<Number>2</Number>
<Reference>101</Reference>
<Number>3</Number>
<Reference>100</Reference>
<Number>4</Number>
<Reference>102</Reference>
<Number>5</Number>
<Reference>100</Reference>
</Root>
预期输出:
<Root>
<Number>1</Number>
<Reference>100</Reference>
<RefNumber>1</RefNumber>
<Number>2</Number>
<Reference>101</Reference>
<RefNumber>1</RefNumber>
<Number>3</Number>
<Reference>100</Reference>
<RefNumber>2</RefNumber>
<Number>4</Number>
<Reference>102</Reference>
<RefNumber>1</RefNumber>
<Number>5</Number>
<Reference>100</Reference>
<RefNumber>3</RefNumber>
</Root>
如何在xslt 1.0中基于根/引用进行分组并将序号添加到输出中的RefNumber?
提前感谢
实现这一点的一种方法是使用xsl:number。每当您匹配Reference元素时,复制该元素,并添加一个RefNumber的元素,该元素具有相同值的Reference元素数:
<xsl:template match="Reference">
<xsl:copy-of select="." />
<xsl:variable name="Ref" select="." />
<RefNumber><xsl:number count="Reference[. = $Ref]" /></RefNumber>
</xsl:template>
这是完整的XSLT
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="Reference">
<xsl:copy-of select="." />
<xsl:variable name="Ref" select="." />
<RefNumber><xsl:number count="Reference[. = $Ref]" /></RefNumber>
</xsl:template>
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
当应用于示例XML时,将输出以下内容:
<Root>
<Number>1</Number>
<Reference>100</Reference>
<RefNumber>1</RefNumber>
<Number>2</Number>
<Reference>101</Reference>
<RefNumber>1</RefNumber>
<Number>3</Number>
<Reference>100</Reference>
<RefNumber>2</RefNumber>
<Number>4</Number>
<Reference>102</Reference>
<RefNumber>1</RefNumber>
<Number>5</Number>
<Reference>100</Reference>
<RefNumber>3</RefNumber>
</Root>
请注意使用标识转换模板来复制其他现有节点。