Hi, for the 5 parents nodes, create a property with the image in it and, in the template, call the image recursively, so the child will look for the parent.

ok, you gotta put that in idiot speak... ROFL.  I am not a guru, I'm a cut and paste kinda guy.

Assuming each parent node is doc type TextPage and template TextPage... both of which are used in child nodes as well...

I'm gonna have to make new doc types I bet... 

Did I mention this site is already live?  LOL

Ok, so it's live... ok, you'll put a macro in your header that will choose the right img.

the macro :

if test="contain(umbraco.library:NiceUrl($currentPage/@id),'parent folder')"

img

/if

And you do that for all you headers. The test says : "if the url contains 'parent folder', then img".

Hope it will help!

System.Xml.Xsl.XslLoadException: 'contain()' is an unknown XSLT function.

    <xsl:if test="contain(umbraco.library:NiceUrl($currentPage/@id),'parent folder')">
      <img src='<xsl:value-of select="PageHeader"/>' />
    </xsl:if>

I'm still alive!! Ok, contains with an S... my bad! You have to replace the parent folder.

Exemple : My parent node is "Students" and, under I have child nodes "Homework" and "Exam". When I'll go to exam.aspx, the url will look like : www.mysite.com/students/exam.aspx

So, the test will check if the word 'students' (whitout the ' ')  is in the url, and, if yes, it will display your img. If not, it will pass is path.

  <xsl:if test="contains(umbraco.library:NiceUrl($currentPage/@id),'parent1')">
    <img src="/image/header1.jpg" />
  </xsl:if>
  <xsl:if test="contains(umbraco.library:NiceUrl($currentPage/@id),'parent2')">
    <img src="/image/header2.jpg" />
  </xsl:if>
...
<img src="/image/default_header.jpg"/>

Is it clearer?

is working on a reply...