Difference between revisions of "Issue:Output nothing if using include in parser function version of DPL within template"

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# If I use <tt>[[debug]]=5</tt>, I can see the non-parsed Wiki text output correctly.
 
# If I use <tt>[[debug]]=5</tt>, I can see the non-parsed Wiki text output correctly.
 
# If I use it individually (not within the template and replace those variable), it also works correctly.
 
# If I use it individually (not within the template and replace those variable), it also works correctly.
 +
# If I remove "[[include]]", it works.
 
# Since I have to pass parameters, I can only use the parser function version.
 
# Since I have to pass parameters, I can only use the parser function version.
  
 
My MediaWiki version: 1.13.0. Is there any clue I can investigate further? Thanks!
 
My MediaWiki version: 1.13.0. Is there any clue I can investigate further? Thanks!
  
[[User:Ericyu|Ericyu]] 09:35, 22 October 2008 (UTC)
+
[[User:Ericyu|Ericyu]] 09:36, 22 October 2008 (UTC)
 +
 
 
== Reply ==
 
== Reply ==

Revision as of 11:36, 22 October 2008

Description: Output nothing if using include in parser function version of DPL within template
Extension / Version: DPL   /   1.7.4
Type / Status: Bug   /   open

Problem

When I use parser function version of DPL within a template, the "include" function will make the whole wiki page output nothing (i.e., empty HTML), and there seems no error in MediaWiki's debug log. I use it like this:

{{#dpl:
|category={{{1}}}
|include=#SomeSection
|titleregexp={{{2}}}
|format=,<h2>'''[[%PAGE%|%PAGE%]]'''</h2>
}}
  1. If I use debug=5, I can see the non-parsed Wiki text output correctly.
  2. If I use it individually (not within the template and replace those variable), it also works correctly.
  3. If I remove "include", it works.
  4. Since I have to pass parameters, I can only use the parser function version.

My MediaWiki version: 1.13.0. Is there any clue I can investigate further? Thanks!

Ericyu 09:36, 22 October 2008 (UTC)

Reply