as the title says I'm trying to create a view that displays rendered files and nodes together. Normally when you create a view you have to select what type of content (nodes, taxonomy, files ...) you want do display.
What i want to achieve is a view that displays all nodes and files. The files are NOT mandatory associated with a node. But they are also tagged with terms.
Any idea how to solve that?
Thanks in advance,
Fab
Why don't you create 2 block views and then display it in a page?
For example:
Create a view block that displays the files --> "display_files-block"
Create a view block that displays the nodes--> "display_nodes-block"
and then print them in a static page u create,
For example: "Display files and nodes", with id 1.
inside page--node--1.tpl.php write:
$block = module_invoke('views', 'block_view', 'display_files-block');
print render($block['content']);
$block = module_invoke('views', 'block_view', 'display_nodes-block');
print render($block['content']);
This doesn't answer the question correctly but can be used as a work around for getting data from two entities to appear in one view.
Through the UI, you could create a view called FilesAndNodes, and only add Global:Custom Text Area under the fields section.
Then create another view called Files, and add this view under the Header section of FilesAndNodes view.
And then do the same for nodes.
The Block created by the FilesAndNodes view can then be added to the page, and the content appears as one view.
Related
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).
I created a view that makes a block for my homepage. I need to render this block in my PHP. Problem is, I need the nid to render the node and I usually get these by going to Structure -> Blocks -> Click configure and the ID is in the page url at that point. For this view I created, there is no ID, just /admin/structure/block/manage/views/cfps-block/configure.
How can I get the ID so I can render this programmatically?
If i understand your problem correctly.
Open you view
For example:
Blockquote
example.com/admin/structure/views/view/custom_view/edit
get the Machine Name of that view suppose it is block, than you can render it by using following code.
echo views_embed_view('custom_view', 'block');
Read the manual if any confusion:
https://api.drupal.org/api/views/views.module/function/views_embed_view/7
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 have some problem with layout in CakePHP. I want have one things whick will be in all pages - login form - but it is in view... How Can I add view(/Users/login.ctp) to layout?
If you want to have some HTML that is repeated among several different views you can use an element.
Create a .ctp file in the "/app/view/Elements" folder. For example, "/app/view/Elements/loginform.ctp".
In that file you would create your login form. Everything available in your view is available in this file (for example the HTML Helper).
To insert this element just do echo $this->element('loginform'); . You can do this both in your view file, or in your layout file.
I'm working on a Drupal site an need to implement the following:
I have created an about us page template (page--about.tpl.php) and a custom content type for the about us page and linked the 2 using suggestions. I also have a custom content type for staff profiles that I need to add on the about us page in a tabbed format.
I can't seem to find a way to get the staff content to display in the about us page. I would ideally like it to render it the same manner as blog posts would display in a blog page.
So my question is, what code do I use to render all the nodes of the staff profiles content type in the about template page?
p.s I'm a bit of a Drupal noob, done a lot of reading but come up empty on this one.
If you already created content type About us (that is what i understood from mockup and explanation) then maybe you could use Views with EVA. That will enable you to have view as field in content type.
You can set it up with manage display or print it in template as all other fields. For example:
<?php print render($content['your_view_entity_view_1']); ?>
Other way would be to embed view in template. For example:
<?php print views_embed_view('your_view', 'block'); ?>
You can use the excellent Views module to create a block to list all your staffs.
Then you need to place the block that you had created with the help of view module in about-us page.
The Views module will give you suggestions to about which template to use.
EDIT: After the op provided the following image.
After you create a view to show all the Team member nodes you could simply print the view in your about us specific page template using following code.
$view = views_get_view('view name');
print $view->render('display_id');
Another option to do the same thing is, make blocks for all the content, viz. The Firm, The Team, Awards, Technonogly, Services and use quicktabs to display the content.
Yet another option to show a view as a field for a node is use EVA
Going the quicktabs way you can provide a lot of flexibility of showing teaser in about us page and leading to details about the same. For example each award can be a node in itself.