Get node position as attribute of a element - xslt

Question

I want to Get node position as attribute of a element. I means I want to get id value like Aut-01,Aut-02 inside context element. I mentioned below Input, output and tried code. I am using XSLT 2.0
Input :
<con-group>
<con c-type="aut">
<name>
<fname>Kent-Dennis</fname>
</name>
</con>
<con c-type="aut">
<name>
<fname>dfr-gvfrt</fname>
</name>
</con>
<con c-type="aut">
<con-id con-id-type="ABC11"
>https://wasq.lk/0000-0002-8551-9535</con-id>
<name>
<fname>Glazier</fname>
</name>
</con>
<con c-type="aut">
<con-id con-id-type="ABC12"
>https://wasq.lk/0000-0002-8551-8535</con-id>
<name>
<fname>Glazier</fname>
</name>
</con>
</con-group>
Output should be :
<link ref="https://orcid.org/0000-0002-8551-9535">
<context type="Aut" id="Aut-01">
<image ref="../../../../command/Templates/Template Art/Auth_.jpg"/>
</context><s/>
<link>
<link ref="https://orcid.org/0000-0002-8551-9535">
<context type="Aut" id="Aut-02">
<image ref="../../../../command/Templates/Template Art/Auth_.jpg"/>
</context><s/>
<link>
Tried code :
<xsl:template match="con-id">
<xsl:text> </xsl:text>
<link ref="{.}">
<xsl:variable name="aaa" select="count(self::con-id/preceding-sibling::*)+1"/>
<context type="Aut" id="{$aaa}">
<image ref="../../../../command/Templates/Template Art/Auth_.jpg"/>
<context><s/>
</link>
</xsl:template>
Help me to solve this. I am getting Id value always 1. As my code I don't get output that I want.

Answer 1

You are getting 1 each time because con-id has no preceding-siblings. (Elements are siblings if they have the same parent element). You should count preceding-siblings of the parent (but only if the preceding siblings have con-id it looks like)
<xsl:variable name="aaa" select="count(parent::*/preceding-sibling::*[con-id])+1"/>
Alternatively, you could use the preceding axis
<xsl:variable name="aaa" select="count(preceding::con-id)+1"/>
You could also use xsl:number here
<xsl:variable name="aaa">
<xsl:number count="con[con-id]" />
</xsl:variable>
If you want to use position() you would have to add another template to match con-group and then only select con-id elements (although this would only really work if you didn't have other processing you wanted to do that might clash).
<xsl:template match="con-group">
<xsl:apply-templates select="*/con-id" />
</xsl:template>
<xsl:template match="con-id">
<xsl:text> </xsl:text>
<link ref="{.}">
<xsl:variable name="aaa" select="position()" />
<context type="Aut" id="{$aaa}">
<image ref="../../../../command/Templates/Template Art/Auth_.jpg"/>
</context><s/>
</link>
</xsl:template>

Answer 2

<?xml version="1.0"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:for-each select="//con-id">
<link ref="{.}">
<xsl:variable name="aaa" select="position()"/>
<context type="Aut" id="{$aaa}">
<image ref="../../../../command/Templates/Template Art/Auth_.jpg"/>
</context>
</link>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>

Answer 3

Check this code:-
<xsl:template match="con">
<xsl:variable name="aaa" select="position()"/>
<xsl:for-each select="con-id">
<xsl:text> </xsl:text>
<link ref="{.}">
<context type="Aut" id="{$aaa}">
<image ref="../../../../command/Templates/Template Art/Auth_.jpg"/>
</context>
</link>
</xsl:for-each>
</xsl:template>