基于条件的XSLT拆分XML文件

Question

首先，我对xslt没有足够的经验。我需要帮助将XML拆分成两个基于InoviceNo的文件。如果InvoiceNo = "+“，则将其包含在前面的兄弟姐妹中，其中InvoiceNo等于一个数字。

源文件

<?xml version="1.0"?>
<ns:MT_CheckoutReport xmlns:ns="http://test.com/LE/ChannelAdvisor">
<row>
    <row>
        <InvoiceNo>547194</InvoiceNo>
        <CheckoutDate>8/02/2021 19:29</CheckoutDate>
        <TotalInvoiceAmt>159.99</TotalInvoiceAmt>
        <PymtType>Amazon</PymtType>
        <PymtID></PymtID>
        <PymtExp>0/0</PymtExp>
        <CVV></CVV> 
    </row>
</row>
<row>
    <row>
        <InvoiceNo>547195</InvoiceNo>
        <CheckoutDate>8/02/2021 19:29</CheckoutDate>
        <TotalInvoiceAmt>219.98</TotalInvoiceAmt>
        <PymtType>Amazon</PymtType>
        <PymtID></PymtID>
        <PymtExp>0/0</PymtExp>
        <CVV></CVV> 
    </row>
</row>
<row>
    <row>
        <InvoiceNo>+</InvoiceNo>
        <CheckoutDate></CheckoutDate>
        <TotalInvoiceAmt></TotalInvoiceAmt>
        <PymtType></PymtType>
        <PymtID></PymtID>
        <PymtExp></PymtExp>
        <CVV></CVV> 
    </row>
</row>
</ns:MT_CheckoutReport>

期望输出档案1

<?xml version="1.0"?>
<ns:MT_CheckoutReport xmlns:ns="http://test.com/LE/ChannelAdvisor">
<row>
    <row>
        <InvoiceNo>547194</InvoiceNo>
        <CheckoutDate>8/02/2021 19:29</CheckoutDate>
        <TotalInvoiceAmt>159.99</TotalInvoiceAmt>
        <PymtType>Amazon</PymtType>
        <PymtID></PymtID>
        <PymtExp>0/0</PymtExp>
        <CVV></CVV> 
    </row>
</row>

文件2. InvoiceNo =+将其包含在InvoiceNo = 547195的兄弟姐妹的文件中。

<?xml version="1.0"?>
<ns:MT_CheckoutReport xmlns:ns="http://test.com/LE/ChannelAdvisor">
<row>
    <row>
        <InvoiceNo>547195</InvoiceNo>
        <CheckoutDate>8/02/2021 19:29</CheckoutDate>
        <TotalInvoiceAmt>219.98</TotalInvoiceAmt>
        <PymtType>Amazon</PymtType>
        <PymtID></PymtID>
        <PymtExp>0/0</PymtExp>
        <CVV></CVV> 
    </row>
</row>
<row>
    <row>
        <InvoiceNo>+</InvoiceNo>
        <CheckoutDate></CheckoutDate>
        <TotalInvoiceAmt></TotalInvoiceAmt>
        <PymtType></PymtType>
        <PymtID></PymtID>
        <PymtExp></PymtExp>
        <CVV></CVV> 
    </row>
</row>

我的尝试

    <xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="row">
  <xsl:for-each-group select="row/InvoiceNo" group-by="InvoiceNo">
    <xsl:result-document href="file_{current-grouping-key()}.xml">
       <some>
         <xsl:copy-of select="current-group()"/>
       </some>
    </xsl:result-document>
  </xsl:for-each-group>
</xsl:template>
</xsl:stylesheet> 

Answer 1

为什么不简单地：

XSLT2.0

<xsl:stylesheet version="2.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ns="http://test.com/LE/ChannelAdvisor">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

<xsl:template match="/ns:MT_CheckoutReport">
    <xsl:for-each-group select="row" group-starting-with="row[row/InvoiceNo!='+']">
        <xsl:result-document href="file_{row/InvoiceNo}.xml">
            <ns:MT_CheckoutReport>
                <xsl:copy-of select="current-group()"/>
            </ns:MT_CheckoutReport>
        </xsl:result-document>
    </xsl:for-each-group>
</xsl:template>

</xsl:stylesheet>