20 Dec
2016
20 Dec
'16
2:47 p.m.
Hi,
I modified the Navigation panel menu in order to have a structured tree menu
that always shows FOO page root and the opened path to reach the current
node (FOO project has nested pages).
{{velocity}}
#panelheader("FOO MENU")
## Escape special characters in macro parameter values.
#set ($openTo = $services.model.serialize($doc.documentReference,
'default').replaceAll('([~"])', '~$1'))
{{documentTree showTranslations="false" showAttachments="false"
compact="true" showRoot="true" openTo="document:$openTo"
root="document:xwiki:FOO.WebHome" /}}
#panelfooter()
{{/velocity}}
Considering all my projects are inside Main\projects page, is there a way to
find dinamically the toplevel page node without having to specify the root
page name (FOO.WebHome)?
Thanks for reply
-Matteo
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21 Dec
21 Dec
11:06 a.m.
On Tue, Dec 20, 2016 at 3:47 PM, matmassa <matmassa(a)eng.it> wrote:
Hi,
I modified the Navigation panel menu in order to have
a structured tree
menu
I would rather create a new panel, because otherwise you may have to
resolve merge conflicts when upgrading XWiki (if the default Navigation
panel changes in any way).
that always shows FOO page root and the opened path to
reach the current
node (FOO project has nested pages).
{{velocity}}
#panelheader("FOO MENU")
## Escape special characters in macro parameter values.
#set ($openTo = $services.model.serialize($doc.documentReference,
'default').replaceAll('([~"])', '~$1'))
{{documentTree showTranslations="false" showAttachments="false"
compact="true" showRoot="true" openTo="document:$openTo"
root="document:xwiki:FOO.WebHome" /}}
#panelfooter()
{{/velocity}}
Considering all my projects are inside Main\projects
page, is there a way
to
This part "inside Main\projects page" is not clear. Can you give an example
of a project reference / URL?
find dinamically the toplevel page node without having
to specify the root
page name (FOO.WebHome)?
All pages that are nested inside FOO have "FOO" as part of their reference:
FOO.A.WebHome
FOO.A.B.WebHome
FOO.X.Y.Z.WebHome
You have the reference of the current document, $doc.documentReference. I
guess you need to find the top level parent of the current document. You
need to use the entity reference API in this case. See
http://extensions.xwiki.org/xwiki/bin/view/Extension/Model+Module
Hope this helps,
Marius
Thanks for reply
-Matteo
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4:37 p.m.
As you suggested I created a copy of the navigation panel to work on.
My xwiki pages are structured as follows:
https://url/xwiki/bin/view/Projects/FOO/
https://url/xwiki/bin/view/Projects/FOO2/
[...]
Inside "Projects" there are several "main" pages for each project (FOO
is an
example of the main page of a project).
FOO page has a tree structure as you correctly wrote:
FOO.A.WebHome
FOO.A.B.WebHome
FOO.X.Y.Z.WebHome
I thought there were be something similar to manage the new pages logic (an
easier way to navigate nested pages).
Your suggestion about creating a new document reference using document
reference works well, I "played" with substrings:
#set ($index = $openTo.indexOf('xwiki:') +6)
#set ($index2 = $openTo.indexOf('.',16))
#set ($subb = $openTo.substring($index,$index2))
#set ($resultpage= $subb.concat('.WebHome'))
Is there a better way to gain this result?
Thank you again,
-Matteo
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5:28 p.m.
On Wed, Dec 21, 2016 at 5:37 PM, matmassa <matmassa(a)eng.it> wrote:
As you suggested I created a copy of the navigation
panel to work on.
My xwiki pages are structured as follows:
https://url/xwiki/bin/view/Projects/FOO/
https://url/xwiki/bin/view/Projects/FOO2/
[...]
Inside "Projects" there are several "main" pages for each project
(FOO is
an
example of the main page of a project).
FOO page has a tree structure as you correctly wrote:
FOO.A.WebHome
FOO.A.B.WebHome
FOO.X.Y.Z.WebHome
I thought there were be something similar to manage the new pages logic (an
easier way to navigate nested pages).
Your suggestion about creating a new document reference using document
reference works well, I "played" with substrings:
#set ($index = $openTo.indexOf('xwiki:') +6)
#set ($index2 = $openTo.indexOf('.',16))
#set ($subb = $openTo.substring($index,$index2))
#set ($resultpage= $subb.concat('.WebHome'))
Is there a better way to gain this result?
Yes.
$services.model.resolveDocument('', 'default', $
doc.documentReference.spaceReferences.get(1))
assuming there are always at least 2 level of space references ("Projects"
and the project itself), but you can do some checks to be safe.
Hope this helps,
Marius
Thank you again,
-Matteo
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Marius Dumitru Florea -
matmassa