Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 1023295
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 16, 20262026-05-16T11:35:41+00:00 2026-05-16T11:35:41+00:00

I have some XML that looks something like this: <Root> <Documents> <Document id=1/> </Documents>

  • 0

I have some XML that looks something like this:

<Root>
    <Documents>
        <Document id="1"/>
    </Documents>
    <People>
        <Person id="1"/>
        <Person id="2"/>
    </People>
    <Links>
        <Link personId="1" documentId="1"/>
        <Link personId="1" documentId="1"/>
        <Link personId="2" documentId="1"/>
    </Links>
</Root>

And I am interested in getting only the ‘Link’ elements that have a unique combination of ‘personId’s and ‘documentId’s, so these two links:

<Root>
    <Links>
        <Link personId="1" documentId="1"/>
        <Link personId="2" documentId="1"/>
    </Links>
</Root>

How might I go about doing that? I have found this question, though I feel mine is slightly more complex and may not apply…I presumme I am going to need to use the key() function somewhere…

Thanks in advance.

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-16T11:35:42+00:00Added an answer on May 16, 2026 at 11:35 am

    This stylesheet:

    <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
        <xsl:key name="kDocAndPeoById" match="Document|Person" use="@id"/>
        <xsl:key name="kLinksByIds" match="Link" 
                 use="concat(@personId,'++',@documentId)"/>
        <xsl:template match="node()|@*">
            <xsl:copy>
                <xsl:apply-templates select="node()|@*"/>
            </xsl:copy>
        </xsl:template>
        <xsl:template match="Documents|People|
         Link[count(.|key('kLinksByIds',concat(@personId,'++',@documentId))[1])!=1
              or not(key('kDocAndPeoById',@personId)/self::Person)
              or not(key('kDocAndPeoById',@documentId)/self::Document)]"/>
    </xsl:stylesheet>
    

    Output:

    <Root>
        <Links>
            <Link personId="1" documentId="1"></Link>
            <Link personId="2" documentId="1"></Link>
        </Links>
    </Root>
    

    If you have no interest into checking if there is such Document or Person @id, then this stylesheet:

    <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
        <xsl:key name="kLinksByIds" match="Link" 
                  use="concat(@personId,'++',@documentId)"/>
        <xsl:template match="node()|@*">
            <xsl:copy>
                <xsl:apply-templates select="node()|@*"/>
            </xsl:copy>
        </xsl:template>
        <xsl:template match="Documents|People|
      Link[count(.|key('kLinksByIds',concat(@personId,'++',@documentId))[1])!=1]"/>
    </xsl:stylesheet>
    

    Output:

    <Root>
        <Links>
            <Link personId="1" documentId="1"></Link>
            <Link personId="2" documentId="1"></Link>
        </Links>
    </Root>
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I have a series of xml files that looks something like this: <ROOT> <F
Alright I have an xml document that looks something like this: <xml> <list> <partner>
I have some XML that looks something like this: <ExtensionObject> <Value xmlns=> <Key>key01</Key> <StringValue>somewords</StringValue>
I have some data in an XML element that looks like this: <?xml version=1.0
I have a BizTalk schema that looks something like this: <Root> <Header> <SomeData></SomeData> <Detail>
I have some XML code that looks like this <SEARCHRESULTS> <FUNCTION name=BarGraph> <PARAMETER name=numList></PARAMETER>
I have an xml that looks like this: <Root> <tag1>4</tag1> <tag2>aa</tag2> <tag3 xmlns:xsi=http://www.w3.org/2001/XMLSchema-instance xmlns:xsd=http://www.w3.org/2001/XMLSchema>
I have some data in an XML element that looks like this: <item value=category1,category2>Item
Let's say I have a file-system that looks a little something like this: C:\stuff\build.xml
Imagine I have some XML that looks like this: DECLARE @xml XML SET @xml

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.