Drupal Views block content based on taxonomy page URL

Question

I have a URL example.com/east-midlands which is a list of all nodes tagged with the term East Midlands.

East Midlands is the parent of several child terms (towns and cities).

I want to create a block that will list these child terms (as links) when you are on the above URL.

How would I go about doing this. I'm assuming Views will creep into this but I've hit a brick wall!

Many Thanks,

Steve

To explain a bit further...

The taxonomy vocabulary is:

Region 1
 - Area 1
  - City 1
  - City 2
 - Area 2
  - City 3
  - City 4
Region 2

And so on...

If you are on URL example.com/region-1/area-1 the taxonomy terms displayed in the block should be City 1 and City 2. If you are on URL example.com/region-1/area-2 the taxonomy terms displayed in the block should be City 3 and City 4.

Hope this is clearer.

S

maged adel maged adel · Accepted Answer · 2011-09-04T17:56:21

1 - Enable "views module" and go to views page http://example.com/admin/structure/views

2 - click on "Add new view" ==> insert the name of ur view page ==> click on "Continue & edit"

3 - click on "add" ==> select "block" ==> click on "add" in the "FILTER CRITERIA" Part

4 - select "Content: Has taxonomy term" ==> click on "Add and configure filter criteria"

5 - click "apply and continue" ==> and write the term name in the auto complete field

6 - after apply all settings click save and you will find a new block generated

i hope that will help u and im sorry if i missed something or if i didn't understand u correctrly

Drupal Views block content based on taxonomy page URL

1 Answers