问题描述
我想显示一个 Drupal 视图，而没有通常围绕它的页面模板 - 我只想要视图节点的纯 HTML 内容.
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.
此视图将包含在另一个非 Drupal 站点中.
This view would be included in another, non-Drupal site.
我希望必须对多个视图执行此操作，因此最好的解决方案是让我快速轻松地设置这些视图 - 我不想每次都创建 .tpl.php 文件我需要在某处包含一个视图.
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.
推荐答案
我一直在寻找一种通过 ajax 拉取节点数据的方法，并为 Drupal 6 提出了以下解决方案.实施以下更改后，如果添加 ajax在 URL 中 =1(例如 mysite.com/node/1?ajax=1)，您将只获得内容，而没有页面布局.
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.
在您的主题的 template.php 文件中:
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';
}
}
然后使用以下内容在您的主题目录中创建 page-ajax.tpl.php:
then create page-ajax.tpl.php in your theme directory with this content:
<?php print $content; ?>
这篇关于显示一个没有页面模板的 Drupal 视图的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！