我必须生成一个xml文件,其中某些分组元素(分组行的文档)用序列编号。反过来,这些组属于另一组(按文档类型)。无论小组如何,编号在所有文档中都必须是连续的。
请参阅示例输出中的Documents元素中的 DocId 属性:
INPUT
<Entries>
<Entry>
<UserID>1</UserID>
<DocNumber>1002</DocNumber>
<Description>An invoice</Description>
<Amount>3103.2000</Amount>
<DocType>INVOICE</DocType>
</Entry>
<Entry>
<UserID>2</UserID>
<DocNumber>3001</DocNumber>
<Description>Some receipt</Description>
<Amount>2352.0000</Amount>
<DocType>RECEIPT</DocType>
</Entry>
<Entry>
<UserID>1</UserID>
<DocNumber>1002</DocNumber>
<Description>An invoice</Description>
<Amount>2861.8400</Amount>
<DocType>INVOICE</DocType>
</Entry>
<Entry>
<UserID>2</UserID>
<DocNumber>3001</DocNumber>
<Description>Some receipt</Description>
<Amount>2352.0000</Amount>
<DocType>RECEIPT</DocType>
</Entry>
<Entry>
<UserID>5</UserID>
<DocNumber>1004</DocNumber>
<Description>Another invoice</Description>
<Amount>2.34</Amount>
<DocType>INVOICE</DocType>
</Entry>
</Entries>
XSLT 2.0
<xsl:stylesheet exclude-result-prefixes="xs xdt err fn" version="2.0" xmlns:err="http://www.w3.org/2005/xqt-errors" xmlns:fn="http://www.w3.org/2005/xpath-functions" xmlns:xdt="http://www.w3.org/2005/xpath-datatypes" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:ser="http://webservice">
<xsl:output indent="yes" method="xml"/>
<xsl:template match="/Entries">
<ser:Request>
<xsl:for-each-group select="Entry" group-by="DocType">
<xsl:sort select="current-grouping-key()"/>
<ser:ItemFile ImportName="{DocType}" Date="{current-date()}">
<xsl:for-each-group select="current-group()" group-by="DocNumber">
<xsl:sort select="current-grouping-key()"/>
<ser:Documents DocId="{position()}" DocRef="{DocNumber}" Desc="{Description}" >
<xsl:for-each select="current-group()">
<Line amount="{Amount}"/>
</xsl:for-each>
</ser:Documents>
</xsl:for-each-group>
</ser:ItemFile>
</xsl:for-each-group>
</ser:Request>
</xsl:template>
</xsl:stylesheet>
输出
<ser:Request xmlns:ser="http://webservice">
<ser:ItemFile ImportName="INVOICE" Date="2017-10-13+02:00">
<ser:Documents DocId="1" DocRef="1002" Desc="An invoice">
<Line amount="3103.2000"/>
<Line amount="2861.8400"/>
</ser:Documents>
<ser:Documents DocId="2" DocRef="1004" Desc="Another invoice">
<Line amount="2.34"/>
</ser:Documents>
</ser:ItemFile>
<ser:ItemFile ImportName="RECEIPT" Date="2017-10-13+02:00">
<ser:Documents DocId="1" DocRef="3001" Desc="Some receipt">
<Line amount="2352.0000"/>
<Line amount="2352.0000"/>
</ser:Documents>
</ser:ItemFile>
</ser:Request>
在此示例中,最后一个DocId属性的所需输出将为序列后的3。
我注意到如果我使用position()函数,编号会在每个组中重新开始,这不是我想要的。我也尝试过xsl:number元素但没有成功。我考虑计算第一组中的元素并将其添加到position(),这对于我来说对于有限数量的组来说是有效的,但是对于可变数量的组通常不能这样做。
是否有一些相对简单的方法来实现这一目标?提前谢谢。
答案 0 :(得分:0)
我会将新创建的元素存储在变量中,然后将其推送到复制除DocId
属性之外的所有内容的模板,然后您可以使用xsl:number
进行设置:
<xsl:stylesheet exclude-result-prefixes="xs xdt err fn" version="2.0" xmlns:err="http://www.w3.org/2005/xqt-errors" xmlns:fn="http://www.w3.org/2005/xpath-functions" xmlns:xdt="http://www.w3.org/2005/xpath-datatypes" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:ser="http://webservice">
<xsl:output indent="yes" method="xml"/>
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/Entries">
<ser:Request>
<xsl:variable name="result">
<xsl:for-each-group select="Entry" group-by="DocType">
<xsl:sort select="current-grouping-key()"/>
<ser:ItemFile ImportName="{DocType}" Date="{current-date()}">
<xsl:for-each-group select="current-group()" group-by="DocNumber">
<xsl:sort select="current-grouping-key()"/>
<ser:Documents DocId="" DocRef="{DocNumber}" Desc="{Description}" >
<xsl:for-each select="current-group()">
<Line amount="{Amount}"/>
</xsl:for-each>
</ser:Documents>
</xsl:for-each-group>
</ser:ItemFile>
</xsl:for-each-group>
</xsl:variable>
<xsl:apply-templates select="$result"/>
</ser:Request>
</xsl:template>
<xsl:template match="ser:Documents/@DocId">
<xsl:attribute name="{name()}">
<xsl:number select=".." level="any"/>
</xsl:attribute>
</xsl:template>
</xsl:stylesheet>