Posted by george@dynapres.nl on February 15, 2007 at 8:46pm
Jump to:
| Project: | Refine by taxonomy |
| Version: | master |
| Component: | Code |
| Category: | bug report |
| Priority: | normal |
| Assigned: | Unassigned |
| Status: | postponed (maintainer needs more info) |
Issue Summary
This issue is related to http://drupal.org/node/103422
There is no point in displaying a refine block when there is only one item to refine. The supplied patch fixes this.
| Attachment | Size |
|---|---|
| refine_by_taxo-more1term.patch | 746 bytes |
Comments
#1
I am not entirely sure about this. AFAIK there are quite a few cases where one would still want to be able to say 'also in'. Consider all the possible 'refinements' methods, please.
#2
Well, yes, the check only applies if you are only using the AND relation which IMO is the most commonly used when 'refining' (narrowing down your selection).
#3
I am not happy with this change. Maybe I am understanding some logic wrong, but I certainly see no need to ignore the block with only one item. That item still is a useful filter, not?
I would agree that the filter blocks make no sense when there is only one *node* to 'list', but that is not what this patch does, afaiks.
#4
The patch is not complete. But let me try to explain the situation which led me to suggest it. Say you have two categories:
- category "cat1" with term "taxonomy/term/11", "taxonomy/term/12"
- category "cat2" with term "taxonomy/term/21"
and the nodes:
- node "A" associated with "term/11" and "term/21"
- node "B" associated with "term/11" and "term/21"
- node "C" associated with "term/12" and "term/21"
and two refinement blocks "refine with AND interconnected through nodes".
When you select "taxonomy/term/11" you'll see two nodes "A" and "B" and a refinement block pointing to "taxonomy/term/11,21". This selection shows the same nodes "A" and "B" and is useless. Selecting "taxonomy/term/21" will show three nodes and two choices pointing to "taxonomy/term/21,11" and "taxonomy/term/21,12" which is useful.
#5
Yes, I think it makes sense. But then we must calculate the remaining nodes, and not the remaining terms.