Views Content Pane "Related content" by taxonomy - drupal-7

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.

Related

Drupal 7: How do I extract specific field in a taxonomy page

How do I extract specific field for display in a taxonomy page?
I have a custom content type called "film" and each film has a Term Reference field called "casting". As expected I can click on a "casting" (tag) it brings me a page where all films are listed wherever this tag is associated. For expample if I click on "Kate Winslet" from movie Titanic, I land on a page http://localhost/mysite/tags/kate-winslet where other movies of Kate Winslet are listed. Up to this point everything is just fine.
I do not want Drupal to pull in and show default fields like just Title and Body in its own display format. Rather I want it so that I can display a photo from each film, year of release and of course the title and trimmed version of the body. I only want to customize the content of this page so that I have the control over What to Show and Where To Show a specific field value.
This is what I tried:
I cloned and put page.tpl.php in my theme's template folder. Renamed it as page--vocabulary--tags.tpl.php. Then I took out the following line of code (<?php print render($page['content']);?>) from my page--vocabulary--tags.tpl.php. The intention was to check whether the overridden template is actually being accessed by Drupal or not. It does!
But I am not been able to extract fields like field_photo or field_release_date from $page['content]. To get an idea about defined variables and how they are placed I used the following line of code:
<pre><?php /*print var_export(get_defined_vars(), TRUE);*/ ?></pre>. But even from there I could not extract a particular field like I mentioned above. The fields look to be somewhere inside $page['content']['system_main']['nodes'], but I don't know how to get to a specific field directly.
I also created a template.php with the following preprocess hook function:
<?php
function introduction_preprocess_page(&$vars) {
if (arg(0) == 'taxonomy' && arg(1) == 'term' && is_numeric(arg(2))) {
$term = taxonomy_term_load(arg(2));
$vars['theme_hook_suggestions'][] = 'page__vocabulary__' . $term->vocabulary_machine_name;
$vars['content'] = $vars['page']['content']['system_main']['nodes'];
}
}
?>
Both <?php print render($content) ?> and <?php print render($page['content']) ?> print the same result but I want something like <?php render($content['photo_field'])?> which I am not been able to.
I am sorry for making this too long. I have just stepped into Drupal. So wanted to make sure that what I am trying to explain matches exactly what I want to accomplish.
You are probably trying the long way to this.
You can use Views module. It allows to create custom listings querying the database, but also override existent ones, like the case of the taxonomy term page listing.
Once you have the module installed (if it's not yet), particularly the Views UI module, go to /admin/structure/views and scroll to bottom, where disabled views (grayed rows) are. You'll find one called Taxonomy term, described as 'A view to emulate Drupal core's handling of taxonomy/term.'
Click Enable on the right of it and then go to the same place where the Enable link is, click the arrow to unfold and choose Edit.
Once you're in the view edit page, you can manipulate the listing at your convenience, adding/removing fields or whatever you want to do in your particular case. If you are not familiar with Views, I recommend you to learn about it, there is a lot of related content on the web and it is close to essential for Drupal development.
Also, if you want to add more customisation to the page, you can use the same approach with the Panels module, who allows to override system pages (not just listings like Views).

Changing default "Taxonomy term" view to add a filter and change path in Drupal 7

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.

In Drupal 7 views how to give external link (another website) to a title?

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.

Addressing the tab menu in view when using taxonomy

I use taxonomy terms to address my menus(these taxonomy terms show the structure of my site) so when I create a view I should use the taxonomy path (instead of alias) for the page setting (the only way that I can link the view to my taxonomy address path as far as I know :P ). whenever I see the page URL I can see the alias. now I'm trying to set up a page that has 2 tabs. I have set it up but the problem is when I click in the tab I see the taxonomy path instead of alias ? (for example taxonomy/term/6/digital instead of CCTV/digital However default menu tab still work correctly and show the alias CCTV)
Is there any way to show the alias instead of taxonomy term address ?
I assume that you have a view that is filtering your taxonomy terms. If you render the results as fields, each field can have it's formatter. If you need a custom form of the output, you can define your own formatter or generate one via GUI with help of: custom formatters.
In case that you render your output as rendered entity, you need to find the right template file and use the drupal_get_path_alias() function within it.
As last I would hint you a great module that also might help you: taxonomy menu

Drupal 7 - How to create a contextual filter based on an aliased url

CASE:
I've created a content type 'Attorney', and have set a url alias pattern for all attorneys to be 'attorneys/[node:title]'. I'd like to create a view that uses the aliased path to display the information about the attorney. This view should have a 'page' display.
EXAMPLE:
When a user visits 'http://mydomain.com/attorneys/aaron-silber' the view returns data for the Attorney with the name Aaron Silber.
BACKGROUND:
I've searched high and low for a solution to this but can't seem to find one that works for me. Typically I'm asked to create a page view with a url of 'attorneys/%' and add a contextual filter with 'Content: Nid', choosing to provide a default value (type: Raw value from URL, path component 2).
The resources on the web for this case are aweful at best. Lets try to fix it here once and for all.
Thanks!
Use a Views block instead with the visibility set to "Only the listed pages" and supply "attorneys/*" in the textarea. Set the block to display in the main content region. Use the Content: Nid filter with a default value of path component 2, as you were previously attempting.
You can't have an attorney node page and a Views page occupying the same URL.

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