Iterating through Drupal Views inside my template using foreach - php

I'm trying to access a Drupal View through my template and want to iterate through it's results to show data. Something simple, like:
foreach($fields as $field) {
echo $field['name_of_the_field'];
}
So i can have full control of my view display inside my teplate. And, of course, being easier to build my template in Drupal. Any help? Thanks in advance.

You can export a views template and then access what you need through that.
Go to the views, and under the 'Basic Settings' section. Click on the 'Theme: information' link.
Then, you'll see a list possible template links.
Click on one of those links and you'll see textarea with the template. Copy the template to a file with the same name as the link name and put it in your themes folder.
Anyway, once you have the proper template, you'll see what object you can iterate over.

Related

How to modify magento front end pages

I am trying to create a module which has both frontend and backend functionality. Like I need to ask for the city in the home page when the store loads. And all the available cities are entered/managed in backend admin panel.
Before I used to write for only backend things, frontend seems little confusing.
There is a design folder which is completely for theme development.
All the example are little different(https://www.mageplaza.com/magento-2-module-development/,http://inchoo.net/magento-2/how-to-create-a-basic-module-in-magento-2/]2), they have routes.xml, where route_id, and all are defined, but here I don't need any extra route. Need some additional tweaks in frontend pages.
I created module V_name/M_name/adminhtml/block controllers etc view ...
Guide me how to create a module, which has both front end and backend connection, cities should be entered in admin, they should show on the frontend homepage.
For now, I only managed to edit home page content CMS page by adding some HTML which shows a popup with a dropdown for cities when the page loads.
Since you already have the back-end figured out I will focus on front-end. Also, since all you need to do is populate a list that you already have created this should be easy. I did something like this before and I found it easier to just use JSON to query a list of, in your case the cities, and populate the drop down. I don't believe this is the most 'MVP/proper' way to go, but it is less work then the other ways. (At least for me it is. I always prefer the JavaScript option since it allows for easy future page customization.)
To use the JSON method you need to create a Block with a method like the one below. You will see that you will also have to create a Resource Model (I'm not going to go over creating the Resource Model or the details of Blocks since there are much better resources than me already online that will go into every single detail you need.). Once this is complete you can access the data straight from the .phtml page in an easy to use JSON array.
First you need to make sure you are now structuring your Modules properly. The new Block below should be in a structure like this...
app/code/<VENDOR>/<MODULE>/Block/Wrapper.php (or whatever you name it)
The admin Blocks should be in the structure below, which it sounds like you are already know how to do.
app/code/<VENDOR>/<MODULE>/Block/Adminhtml
Create your Block and add a method to create a JOSN array like below...
public function getCityList()
{
$city_array = array();
/** #var \<VENDOR>\<MODULE>\Model\ResourceModel\City\Collection $collection */
$collection = $this->_cityCollectionFactory->create();
$collection->addFieldToFilter('active','1')->addFieldToSelect(['city_id', 'city']);
$collection->getSelect()->order(array('city ASC', 'city_id ASC'));
$count = 0;
foreach ($collection as $model)
{
$city_array["$count"] = $model->getData();
$count++;
}
return \Zend_Json::encode($city_array);
}
FYI... The foreach loop in the code above is weird and uses $count because I needed to do some tricky things to get something to work.
Then you can create the Block in your .phtml file to access the data via javascript.
<?php
$block_obj = $block->getLayout()->createBlock('<VENDOR>\<MODULE>\Block\Wrapper');
?>
<script type="text/javascript">
window.citylistJson = <?php echo $block_obj->getCityList() ?>;
</script>

Overriding an article output in Drupal 7 using a template.php theme

I basically want to modify the output of my articles, by putting a between each article that is listed on my page.
I have overridden other functions in my themes template.php as follows
function mytheme_preprocess_html(&$variables) {
drupal_add_css('http://fonts.googleapis.com/css?family=Gudea', array('type' => 'external'));
}
I am looking for something similar for the articles?
Copy your theme's node.tpl.php to a file called node--article.tpl.php and add an <hr> at the bottom of the file.
Have you looked at the Views module? Take a look http://drupal.org/project/views and use it to list content for you. Views can create a Page display for this content. You can then apply styles to the Views-generated HTML that will allow you to put dividing borders between the nodes listed. Views provides 'first' and 'last' classes as well so that you will not apply a border to the last element.

Theming and layout in yii framework

I am a newbie in Yii Framework and creating a CRM which is module based.
Using different tutorials I am able to create my own theme, but now I am stucked at one point.
In my theme, the upper <nav> and left <nav> remains the same throughout the app, until user is logged in. That's why I made it a part of my main.php, but in the login page there are no buttons to show, just simple login form with 2 textfields.
How can I implement this form in my application using custom themes?
I have tried to define a layout in that particular action but not succeeded. Any help would be appreciated.
Using a custom layout for your view is the right way to go.
You can either set the layout in the controller action or in the view.
$this->layout = "//layouts/mylayout";
Note that the default layouts column1.php and column2.php also use the main.php layout file.
Try this step by step :
Create New theme
You can create a new theme and add this to the directory
Application_Root/themes.
Look at the themes/classic directory to get an an idea of the structure of the directory.
The important file (at this stage) is :-
Application_Root/themes/views/layouts/main.php
Customise your theme contents
Copy the css, image, js files etc to the correct directory and change the main.php file to your liking. For example, if your main.php says
<link href="css/mystyle.css" rel="stylesheet">
Then you will have a file
Application_Root/css/mystyle.css
Create the content placeholder.
Somewhere in your main.php, there will be a placeholder for dynamic text, which is specified by.
<?php echo $content; ?>
Tell yii to use the theme.
Change the file Application_Root/protected/config/main.php by adding the following line just before the last line (containing the closing bracket).
'theme'=>'surveyhub'
Create the layout placeholders.
Create an HTML segment that will be written into the $contents portion of main.php. Call it for example one_column.php. The file path will therefore be Application_Root/themes/views/layouts/one_column.php In that file, where you want the dynamic text to be placed, create a placeholder.
<?php echo $content; ?>
Tell Yii to use the layout.
In the file Application_Root/protected/components/Controller.php, add or modify the layout variable to read :
public $layout='//layouts/one_column.php';
Refresh the page

Drupal 7 theming: region error

this is in my theme.info file of my Drupal theme:
regions[conf_modules] = Configurator: modules
I'm using multiple templates for different node types. In one of them I'd like this region/block to show up. So I've put this in node--configurator.tpl.php:
<?php print render($page['conf_modules']); ?>
In the Drupal administration panel I have assigned a views block to the region, but on the node--configurator.tpl.php pages, the views block is not shown. Am I using render() properly? What's going wrong here? Thanks in advance!
In node templates, $page is simply a status variable that is:
True if the node is being displayed by itself as a page.
However, you can add regions to the page for specific content types through your page.tpl.php if you like. Something like below should work if placed in page.tpl.php:
<?php
if ($node->type == 'CONTENT_TYPE') {
print render($page['conf_modules']);
}
?>

displaying a Drupal view without a page template around it

I would like to display a Drupal view without the page template that normally surrounds it - I want just the plain HTML content of the view's nodes.
This view would be included in another, non-Drupal site.
I expect to have to do this with a number of views, so a solution that lets me set these up rapidly and easily would be the best - I'd prefer not to have to create a .tpl.php file every time I need to include a view somewhere.
I was looking for a way to pull node data via ajax and came up with the following solution for Drupal 6. After implementing the changes below, if you add ajax=1 in the URL (e.g. mysite.com/node/1?ajax=1), you'll get just the content and no page layout.
in the template.php file for your theme:
function phptemplate_preprocess_page(&$vars) {
if ( isset($_GET['ajax']) && $_GET['ajax'] == 1 ) {
$vars['template_file'] = 'page-ajax';
}
}
then create page-ajax.tpl.php in your theme directory with this content:
<?php print $content; ?>
Based on the answer of Ufonion Labs I was able to completely remove all the HTML output around the page content in Drupal 7 by implementing both hook_preprocess_page and hook_preprocess_html in my themes template.php, like this:
function MY_THEME_preprocess_page(&$variables) {
if (isset($_GET['response_type']) && $_GET['response_type'] == 'embed') {
$variables['theme_hook_suggestions'][] = 'page__embed';
}
}
function MY_THEME_preprocess_html(&$variables) {
if (isset($_GET['response_type']) && $_GET['response_type'] == 'embed') {
$variables['theme_hook_suggestions'][] = 'html__embed';
}
}
Then I added two templates to my theme: html--embed.tpl.php:
<?php print $page; ?>
and page--embed.tpl.php:
<?php print render($page['content']); ?>
Now when I open a node page, such as http://example.com/node/3, I see the complete page as usual, but when I add the response_type parameter, such as http://example.com/node/3?response_type=embed, I only get the <div> with the page contents so it can be embedded in another page.
I know this question has already been answered, but I wanted to add my own solution which uses elements of Philadelphia Web Design's (PWD) answer and uses hook_theme_registry_alter, as suggested by Owen. Using this solution, you can load the template directly from a custom module.
First, I added raw.tpl.php to a newly created 'templates' folder inside my module. The contents of raw.tpl.php are identical to PWD's page-ajax.tpl.php:
<?php print $content; ?>
Next, I implemented hook_preprocess_page in my module in the same fashion as PWD (except that I modified the $_GET parameter and updated the template file reference:
function MY_MODULE_NAME_preprocess_page(&$vars) {
if ( isset($_GET['raw']) && $_GET['raw'] == 1 ) {
$vars['template_file'] = 'raw';
}
}
Finally, I implemented hook_theme_registry_alter to add my module's 'templates' directory to the theme registry (based on http://drupal.org/node/1105922#comment-4265700):
function MY_MODULE_NAME_theme_registry_alter(&$theme_registry) {
$modulepath = drupal_get_path('module','MY_MODULE_NAME');
array_unshift($theme_registry['page']['theme paths'], $modulepath.'/templates');
}
Now, when I add ?raw=1 to the view's URL path, it will use the specified template inside my module.
For others who may hit this page, if you're just working with standard callbacks (not necessarily views), this is easy. In your callback function, instead of returning the code to render within the page, use the 'print' function.
For example:
function mymodule_do_ajax($node)
{
$rval = <<<RVAL
<table>
<th>
<td>Data</td>
<td>Data</td>
<td>Data</td>
</th>
<tr>
<td>Cool</td>
<td>Cool</td>
<td>Cool</td>
</tr>
</table>
RVAL;
//return $rval; Nope! Will render via the templating engine.
print $rval; //Much better. No wrapper.
}
Cheers!
Another way to do it which I find very handy is to add a menu item with a page callback function that doesn't return a string:
Example:
/**
* Implementation of hook_menu.
*/
function test_menu(){
$items['test'] = array (
/* [...] */
'page callback' => 'test_callback',
/* [...] */
);
return $items;
}
function test_callback() {
// echo or print whatever you want
// embed views if you want
// DO NOT RETURN A STRING
return TRUE;
}
-- Update
It would be much better to use exit(); instead of return TRUE; (see comment).
Hey, here's yet another way of doing it:
1) Download and install Views Bonus Pack (http://drupal.org/project/views_bonus)
2) Create a Views display "Feed" and use style "XML" (or something you think fits your needs better).
3) If you're not satisfied with the standard XML output, you can change it by adjusting the template for the view. Check the "theme" settings to get suggestions for alternative template names for this specific view (so you'll still have the default XML output left for future use).
Good luck!
//Johan Falk, NodeOne, Sweden
Based on answer of Philadelphia Web Design (thanks) and some googling (http://drupal.org/node/957250) here is what worked for me in Drupal 7 to get chosen pages displayed without the template:
function pixture_reloaded_preprocess_page(&$vars)
{
if ( isset($_GET['vlozeno']) && $_GET['vlozeno'] == 1 ) {
$vars['theme_hook_suggestions'][] = 'page__vlozeno';
}
}
instead of phptemplate, in D7 there has to be the name_of_your_theme in the name of the function. Also, I had to put two underscores __ in the php variable with the file name, but the actual template file name needs two dashes --
content of page--vlozeno.tpl.php :
<?php print render($page['content']); ?>
The output, however, still has got a lot of wrapping and theme's CSS references. Not sure how to output totally unthemed data...
Assuming you're in Drupal 6, the easiest way to do this is to put a phptemplate_views_view_unformatted_VIEWNAME call in template.php (assumes your view is unformatted - if it's something else, a list say, use the appropriate theme function). Theme the view results in this theme call then, instead of returning the results as you normally would, print them and return NULL. This will output the HTML directly.
PS - make sure to clear your cache (at /admin/settings/performance) to see this work.
there are probably a number of ways around this, however, the "easiest" may be just setting your own custom theme, and having the page.tpl.php just be empty, or some random divs
// page.tpl.php
<div id="page"><?php print $content ?></div>
this method would basically just allow node.tpl.php to show (or any of drupal's form views, etc...) and would be an easy way to avoid modifying core, or having to alter the theme registry to avoid displaying page.tpl.php in the first place.
edit: see comments
ok i played around with views a bit, it looks like it takes over and constructs it's own "node.tpl.php" (in a sense) for display within "page.tpl.php". on first glance, my gut feeling would be to hook into theme_registry_alter().
when you're looking at a views page, you have access to piles of information here, as well as the page.tpl.php paths/files. as such i would do something like:
function modulejustforalteration_theme_registry_alter(&$variables) {
if (isset($variables['views_ui_list_views']) ) {
// not sure if that's the best index to test for "views" but i imagine it'll work
// as well as others
$variables['page']['template'] = 'override_page';
}
}
this should allow you to use a "override_page.tpl.php" template in your current theme in which you can remove anything you want (as my first answer above).
a few things:
as i said, not sure if views_ui_list_views is always available to check against, but it sounds like it should be set if we're looking at a view
you can alter the theme paths of the page array if you prefer (to change the location of where drupal will look for page.tpl.php, instead of renaming it altogether)
there doesn't appear to be any identifiers for this specific view, so this method might be an "all views will be stripped" approach. if you need to strip the page.tpl.php for a specific view only, perhaps hooking into template_preprocess_page() might be a better idea.
I like the Drupal module. BUt, here's another way.
copy page.tpl.php in your theme folder to a new file called page-VIEWNAME.tpl.php, where VIEWNAME is the machine-readible name of the view.
Then edit page-VIEWNAME.tpl.php to suit.
There is also http://drupal.org/project/pagearray which is a general solution...
Also, #Scott Evernden's solution is a cross site scripting (XSS) security hole. Don't do that. Read the documentation on drupal.org about how to Handle Text in a Secure Fashion http://drupal.org/node/28984
A simple way to display content of a special content-type you wish to display without all the stuff of the page.tpl.php:
Add the following snippet to your template.php file:
function mytheme_preprocess_page(&$vars) {
if ($vars['node'] && arg(2) != 'edit') {
$vars['template_files'][] = 'page-nodetype-'. $vars['node']->type;
}
}
Add a page-nodetype-examplecontenttype.tpl.php to your theme, like your page.tpl.php but without the stuff you don't want to display and with print $content in the body.
If I understand your question, you want to have nodes which contain all the HTML for a page, from DOCTYPE to </HTML>. What I would do is create a content type for those nodes -- "fullhtml" as its machine-readable name -- and then create a node template for it called node-fullhtml.tpl.php. You can't just dump the node's contents, as they've been HTML-sanitized. node.fullhtml.tpl.php would literally be just this:
echo htmlspecialchars_decode($content);
Then you'll need a way to override the standard page.tpl.php. I think what you could do is at the top of your page.tpl.php check the $node's content type, and bail out if it's fullhtml. Or, set a global variable in node-fullhtml.tpl.php that page.tpl.php would check for.
I'm no Drupal expert, but that's how I'd do it. I'm talking off the cuff, so watch for devils in the details.
I see you have already gone and made yourself a module, so this may no longer help, but it is fairly easy to get a view to expose an rss feed, which might be an easier way of getting at the content, especially if you want to include it on a different site.
On D7 you can use menu_execute_active_handler
$build = menu_execute_active_handler('user', FALSE);
return render($build);
jeroen's answer was what did for me after playing with it. I have a Drupal 7 site.
First of all make sure you replace MY_THEME with your theme name. Yes it is obvious but most newbies miss this.
I actually already had a function MY_THEME_preprocess_page(&$variables) {. Do not recreate the function then but add this code at the end of the function before you close it with }.
if (isset($_GET['response_type']) && $_GET['response_type'] == 'embed') {
$variables['theme_hook_suggestions'][] = 'page__embed';
}
My function used $vars not $variables, so I had to update that as well. Again obvious if you think look for it.
My first answered allowed me to only display the node when I called it up in a web browser. However the ultimate goal of this is to embed the drupal node in an 3rd party site using iframe.
Since the release of Drupal Core 7.50 iframe is by default blocked to prevent Clickjacking
To get only the node to successfully embed in a 3rd party site you also need to override the x-frame default setting. Everything started working after I added the following in template.php
function MY_THEME_page_alter($page) {
if (isset($_GET['response_type']) && $_GET['response_type'] == 'embed') {
header_remove('X-Frame-Options');
}
}

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