I have a custom (i.e. View) menu based on taxonomy vocabulary, in wich each term has an image field.
So far, each term is displayed properly with its picture, but I want to know if it is possible to display only the Image field relative to currently active term, while maintaining label+link for the whole vocabulary. Result would be to have an illustrative picture in top of menu, changing according to which item is clicker.
Is there any way to do something like that directly inside the View configuration ?
With the use of contextual filter, you can display information for the term passed as argument/ in URL.
Related
I have a paragraph named "Car Slider" and a Content Typ "Cars".
The content type consists of different fields like name, make etc. as well as a "Tags" taxonomy reference.
The car slider paragraph consists of fields like headline, description, and a Views Reference to display cars in a slider.
Right now I have different view displays to display only cars with specific tags in the slider, but it sounds unnecessary to have tons of displays, instead I'd prefer to have a dropdown in the backend editor the praragraph where I select the relevant tags and the view filters results to only that tag.
I want to use this to display sliders consisting cars of different categories/tags.
I am aware of the Expose Filter option, but this exposes the filter to the frontend/end-user, but I want to expose it to the admin who decides which content slider to add to the page.
I couldn't find a way to achieve this so far. Can anyone tell me if this is possible and how to achieve this? Thank you!
I'm making a views slideshow of some images.
These images have a title.
I would like to expose this title as a drop-down list filter.
I'm able to do this with pr-configured filter values, but i want this list to be build by the values that the end users gives as a title to a image. (or made of another field like description etc...)
So, when a user uploads images with a title "vacation", i would like this value to come up in the exposed filter in the views slideshow...
Thank you
This can easily be done using the Views Reference Filter module.
Create a views page with the fields you wish to display.
Add an Entity Reference display to your view.
Set which field should act as your search field in the format settings of the Entity Reference display (in this case Content: Title). Remove or exclude all other fields for this display only (override).
To your page display add a filter Content: Nid (entityreference filter) and expose it to visitors. Remember to select the correct view in the "View used to select the entities" drop down if you have more than one Entity Reference views.
Optional steps:
Enable Ajax for your view.
Set the exposed form style setting to auto submit.
If I were trying to accomplish this I would add a taxonomy vocabulary for these images, then I would add a field to the image content type referencing this vocabulary.
After that, I would use create a Rule that would react to saving a new image and add a new taxonomy term to that image that is the title entered.
You can then easily create a filter in views to filter off that taxonomy term and expose that as a dropdown. By doing this each image should have a taxonomy term attached to it that is the image's title. As images are added the taxonomy terms should appear in the dropdown.
Let me know if you need any additional guidance.
I have a menu block to the right of my content area. I want to be able to click on one of the links and have the content in the content area change, whether that be to advance a nodeque to the matching content type or just diplay a block containing that one content type.
I would usually do this with php, by adding a url variable site.com/data?uid=somenumber to display the data i want on the same page. I don't want to have to create multiple content pages to do this. There has to be a way that the links that are created and displayed in the menu block can just change the content inside the view displayed on the page.
Im a total drupal newb. So any help would be incredibly appreciated.
Thanks in advance.
If your content area is displaying a view, you can definitely do this using views arguments. So, for example, if you want the menu links to vary the view content based on different terms, you can set up your view to expect an argument from the url.
To set it up:
Add a 'Contextual filter' to your view (under the 'Advanced' section)
In my example below, I've added 'region' as my contextual filter, which is just a term reference on my content type
When configuring the contextual filter, set it up such that when the filter is not present in the URL, it just displays all values (that's the default) -- shown in screenshot.
4.
This way, when you first load the page, and no menu item has been selected, all your content will show in the view. Now you'll have to set up your menus to provide the argument, and your view should react appropriately!
Let me know if that works or if you have any questions.
Is there an easily managed way we can assign a specific taxonomy term to a specific instance of a view?
We're using the callouts module for Drupal, and it works great for nodes/pages we have in our site. But we have a handful of Views that are displaying as a page, but we cannot assign a specific taxonomy term to a specific instance of that view.
The view itself is pulling in lists of nodes based off of a taxonomy argument:
Example On the bottom left, we have some callouts, but they're shown randomly because we can’t assign a term to that page to show specific callouts.
There's 7-8 of these pages, and they all use the same View, but just passing different data through the argument. we've tried assigning the callouts we want to the taxonomy that is used to create the view, but that didn't work either.
Without changing the view output (for example, from a page to a block), how can this be accomplished?
Even if you could assign a taxonomy term to a view, I think it would not help in this case. Blocks are not aware of their context at the time they are rendered, so they have to look at the url to retrieve information about the node that's being displayed. The block checks if the page displays a node; if so the node data is loaded. The callouts module also works like that, as is explained on the module page. Since a view is not a node, the block will not render. To make this work, you would have to alter the callouts module I'm afraid.
Let's suppose I have created a view that shows some kind of stories.
But I want to show this view in a left-hand bar — not a link to the view, but the view itself.
How can I connect my new view with a fixed block position?
I want to be able to show real view data in various places on my page.
Is it possible or I am limited only to central area and links to views from menu?
Using Views 2.x for Drupal 6.x it's simple to create a block from a view. Every view has a set of 'default' settings and some number of display settings. A display can be a page, a block, a feed or anything else that creative module authors.
To make a block from your view, you just "Add Display" of type "block", override any settings that you want changed in the block (IE - display less items, just a node title, whatever). You then have a block that you can place like any other Drupal block.
edit: Answer to "Can you limit stories to ones that are tagged"? Sure thing. You just add a filter for Taxonomy terms.
Getting a view to display in a block is easy. Once you have created a view and assigned the view as a block position you simply head to blocks in your admin and you'll see the view.
Just select the block and save it. Simple.
If your view doesn't show, check your filters and that the view fields are correct. Usually if they are not showing it is something simple like selecting content published = yes.
You can create views as pages or blocks. As you're not telling exactly which Drupal/Views version you're working with, all I can tell so far is that, after you create the view, you can tell it to display as a block or as a page.
Then go to the blocks settings and set that to the position you want.