XSLT映射父节点的不同子节点

Question

I have recently started using XSLT. I am trying to get a report of this xml set

One of xml looks like the below one and named (company_a.xml)

<company title="abc" >
<section>
<in-proc uid="HR.xml">
    <details>
        <empname>abcd1</empname>
        <id>xyz1</id>
        <empname>abcd2</empname>
        <id>xyz2</id>
        <empname>abcd3</empname>
        <id>xyz3</id>
    </details>
</in-proc>      
<out-proc uid="manufacturing.xml">
    <title>any</title>
    <details>
        <empname>abcd4</empname>
        <id>xyz4</id>
    </details>
</out-proc>
</section>
</company>

XSL used is

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes"/>
    <xsl:template match="/">
    <xsl:result-document href="Emp_data.csv">
            <xsl:for-each select="//file">
                <xsl:variable name="filename" select="."/>
                <xsl:variable name="fileid" select="substring-before(., '.xml')"/>
                <xsl:for-each select="document($filename)">
                    <xsl:for-each select="company/section/*/details">
                        <xsl:variable name="business" select="local-name(..)"/>
                        <xsl:variable name="dept" select="substring-before(ancestor::*/@uid, '.xml')"/>
                            <xsl:value-of select="concat('&quot;', id , '&quot;,&quot;', empname , '&quot;,&quot;', $fileid, '&quot;,&quot;', $dept, '&quot;,&quot;', $business, '&quot;&#xA;')"/>
                    </xsl:for-each>
                </xsl:for-each>
            </xsl:for-each>
    </xsl:result-document>
    </xsl:template>
</xsl:stylesheet>

is giving

xyz1,abcd1,company_a,HR,in-proc
xyz4,abcd4,company_a,manufacturing,out-proc

Results I am looking for is

xyz1,abcd1,company_a,HR,in-proc
xyz2,abcd2,company_a,HR,in-proc
xyz3,abcd3,company_a,HR,in-proc
xyz4,abcd4,company_a,manufacturing,out-proc

I tried using

<xsl:for-each select="id"> 
    <xsl:value-of select="."/>

but then I am not sure how to fetch empname

I am using file_list(document($filename)) as there are more than 100 files which has millions of lines to look within

Please advice.

Answer 1

If you replace

                <xsl:for-each select="company/section/*/details">
                    <xsl:variable name="business" select="local-name(..)"/>
                    <xsl:variable name="dept" select="substring-before(ancestor::*/@uid, '.xml')"/>
                        <xsl:value-of select="concat('&quot;', id , '&quot;,&quot;', empname , '&quot;,&quot;', $fileid, '&quot;,&quot;', $dept, '&quot;,&quot;', $business, '&quot;&#xA;')"/>
                </xsl:for-each>

with

                <xsl:for-each select="company/section/*/details/empname">
                    <xsl:variable name="business" select="local-name(../..)"/>
                    <xsl:variable name="dept" select="substring-before(ancestor::*/@uid, '.xml')"/>
                        <xsl:value-of select="concat('&quot;', following-sibling::id[1] , '&quot;,&quot;', . , '&quot;,&quot;', $fileid, '&quot;,&quot;', $dept, '&quot;,&quot;', $business, '&quot;&#xA;')"/>
                </xsl:for-each>

then you should get a line for each existing empname .