displaying a Drupal view without a page template around it

问题

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.

回答1:

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; ?>

回答2:

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.

回答3:

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.

回答4:

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!

回答5:

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).

回答6:

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

回答7:

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...

回答8:

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.

回答9:

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.

回答10:

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.

回答11:

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

回答12:

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.

回答13:

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.

回答14:

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.

回答15:

On D7 you can use menu_execute_active_handler

$build = menu_execute_active_handler('user', FALSE);
return render($build);

回答16:

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.

回答17:

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');
 } 
}

来源：https://stackoverflow.com/questions/247991/displaying-a-drupal-view-without-a-page-template-around-it

标签

php

drupal

drupal-views