How to display a Views block based on the current page?

Question

I've created a new content type "banner" (title,description, image, price) that I would like to display on certain pages (limited to parent pages). I'm using the Views module and was looking for a contextual filter or something like that but I'm a noob at Drupal and can't figure out my way.

For example, I have 2 content types: rooms and activities, 2 page listings respectively my rooms and my activities and finally I've created two banners, banner-rooms and banner-activities.

How can I create a Views block which displays the right instance according to the current page ? How can I limit it to the parent page (ie: do not display the banner on rooms/roomA, only on rooms)?

Answer 1

Suppose there are two views

views 1 url is - http://example.com/my-rooms

views 2 url is - http://example.com/my-activities

There are two blocks -

a) banner-rooms

b) banner-activities

On the block configuration page, You need to set the visibility setting of the block for specific pages.

=>Under Visibility setting, Select the tab "Pages"

=>Select the option - "only on the listed pages"

In the text-area, fill the urls e.g- banner-rooms

Answer 2

You should be able to get this to work using the Rules block visibility module, using a Rules Component that is pretty similar to what I described in my answer to "How to display a block if a user does NOT have a selected role?". All you'd have to do is to adapt the Rules condition to make it check if the path of the current page is "rooms".