There are nodes of "product" content type which contains reference field to the "version" taxonomy terms.
User can add new term when he is creating a new node (by auto complete widget) and I need to display nodes related to the newest or "higher" term.
Is this possible without creating custom filter grammatically ?
It is just 2 lines of code , you can place it anywhere in your page.tpl.php or
page--front.tpl.php .
$term = max(taxonomy_get_tree(taxonomy_vocabulary_machine_name_load('version')->vid));
if ($term)
echo l("Latest Version Products", 'taxonomy/term/'.$term->tid);
By placing the above code in your template tpl file , you can find
"Lastest Version Product" link. and selecting that link will take user to the latest term page.
Related
I have a Drupal 7 question. I would like to display certain pieces of information from my current node using a view.
I can use views to display information from other nodes easily by setting filters and other sorting features. That's no problem. However, I want to display information from the node which is currently opened. So, for example, if I'm on a page (node: 117, title: news), I can display all the images from that page, plus captions, by creating a view.
The logic, I suppose, would be to create a view with a filter (or contextual filter) stating that the node has to equal the current node, but I don't know how to do that.
Does anyone have any advice? How do I instruct a view to display information from the node which is currently open?
Thanks!
Edit the view
Add Content: Nid in contextual filter
In contextual filter options
WHEN THE FILTER VALUE IS NOT AVAILABLE
Select "Provide default value"
Type: Content ID from URL
Now the default argument will be the nid of the current node.
I am trying to do a seemingly simple thing but unable to come up with anything.
I'll be specific here for what I want to achieve. I want to clone the existing Taxonomy term which has a path of /taxonomy/term/%. After cloning this view, I wanna make the following changes to it:-
Change the name to Document of Taxonomy Term
Add a filter to this view of a field called is_document. It's a checkbox which is when clicked shows that this node of my taxonomy term is a document.
Change path to /taxonomy/term/documents/%, which shows that I am opening all nodes of this term which are marked as document.
Now when I open my site, let's say with the URL www.mysite.com/organisation/company/documents, it doesn't pick the view. The URL www.mysite.com/organisation/company works as it picks the default Drupal View.
Any idea what I can do here?
I hope I was able to explain efficiently.
UPDATE
I used TVI, but it changes all the pages of my taxonomy terms. I don't wanna change my current scenario where normal taxonomy term pages are opened through /taxonomy/term/% (native Drupal view). I just wanna open all taxonomy terms with a document filter when /taxonomy/term/documents/% view is selected.
Thanks.
I suggest you to download and install Pathauto for URL aliases patterns.
In your Views, you set the path at: taxonomy/term/documents/%
In Pathauto settings, you to go "Patterns", then scroll to the taxonomy section where you can set a new URL pattern: [term:vocabulary]/[term:name]/documents
Then, go to Update tab and update all the paths for Taxonomy.
Enjoy your new Views.
Iam new to drupal 7 views.
I have a content type contains title, description fields. I want the content title's to scroll at the top of the page, so that i created a view and it works fine.What is my question is, i want to link a content (eg: 1st content (title) in a scroll) to another website instead of content page, the remaining contents linked to the content page. Is it possible?.If it is possible how it can be done?..
Thanks in advance,
A.John Melchior.
Yes, it's possible.
One way to do it, would be to use the Link module to create a link field in your node.
Use the link field to input the external link value.
In views add the link field, before the title ( order is important ) and exclude it from display
Add the node title field in the views and in the rewrite field output option, use one the link field token as your path.
I'm not looking at the Views UI right now , but you should find at least a couple of ways to redirect content when the node title is clicked. You should have the Rewrite field output and the Output this field as a link.
You could make use of Display Suite coupled with the description above.
Create a link field for your nodes.
Create 2 view modes for your content type via Display Suite. View Mode 1 will show the link field as the first data element, View Mode 2 will show the title as the first element.
Configure your view to show output using Display Suite, configured to show the first record using View Mode 1, all others using View Mode 2.
I'm currently developing a Drupal 7 site where I use Page Manager's "node_view" to customize the layout for my "Article" content type.
So, whenever viewing an Article I want to show a little box with "Related content".
I have a taxonomy vocabulary "Shared category" that I use in all my content types.
For example I also have a "Gallery" content type, so when viewing an article in with the "myTag" term - I want to show a Content Pane with images tagged with "myTag" from the "Shared category" vocabulary.
My question is about setting up the Contextual Filter for the Content Pane.
Should the "Argument input" be a Vocabulary or a Term? - I've tried both but can't get it to work.
Also I have another question: Is it possible to use a Content Pane as a block, that is shown with the Panel? - I guess not, sinse the Contextual filter value comes from Page Manager.
You won't want to manually input the argument in the panel, as it will vary for each article.
You'll want to set it automatically using a default argument that gets the tid from the node.
To get the related content block to show content based on the same terms as the node being viewed, you'd need to add a taxonomy term id argument to the view.
You can then try the 'Set a default argument' and use 'taxonomy term from URL' ( node option )
If that doesn't work, you may need to get the relevant tid from node using php code in the default argument area, with something like node_load or menu_get_object to load the node.
e.g something like
if($node = menu_get_object()){
if(isset($node->field_shared_category[$node->language][0]['tid'])){
return $node->field_shared_category[$node->language][0]['tid'];
}
}
Related content can be shown by using contextual filter in block views. This block view will contain content which are associated with the current content viewed by a user.
You can goto "Create related content block view based on terms in Drupal 7" blog post to get step by step instruction to create this block view.
Developing my custom Drupal's theme. it will contain custom node.tpl.php file.
How can i get and print related taxonomy names of selected node?
Tnx in adv!
EDIT: Doh, my apologies, I'm just now seeing the Drupal 7 tag, specifically. It appears this thread has some possible solutions: http://drupal.org/node/909968
With D6 (not 100% about D7) In the node's template .php files (and similarly, in a view or most anywhere you have access to a node's properties with custom PHP, like a View or Block), you can use the following:
// returns array of taxonomy objects for given node
$tax_terms = taxonomy_node_get_terms($node);
// prints each term name
foreach ($tax_terms as $tax) {
print $tax->name;
}
Also, there's a few useful Drupal functions for cases like this:
// print_r's all properties of a given node, similar to devel
dpr($node);
// using this in the above 'for' look will give you all properties of each taxonomy object
dpr($tax);
Here's a website that lists a few more of these functions.