I create a content type with fields title header and body and in the page.tpl I want to print in only the header because I have a css that make the header in a box i tried with render($content['header']);
but is not working
ok well i finish with this I made a tpl.php for every node i wanted to edit and in drupal 7 i have to put hidden($content[heder]) or whatever the name of the file in the node is and after print the content for that node I kept the different modifications in variables and I print the variables
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I am a new in Drupal. I want your help for doing this please help
I have created one Content type with fields( Title, Body, File) and also added content through this content type. But Some of the content didn't have file filed and some of the content having the only file. In the view, I want to show both either body or file. if suppose file field is empty then it should display the body of the same content.
Thanks.
You could do this by creating a custom node template for particular content type. So if you have a content type 'news', for example, the you can create a node template for this content type and customized the front-end display of the content. See below how it would work.
As per the Drupal theme suggestions, create a node template, named, node--news.tpl.php (copying the existing node.tpl.php file) in your custom (active) theme directory under the templates folder. On this template, you will be able to access the $node object variable, containing complete node information including fields and data.
Suppose your fields are field_image and body then replace your template's code print render($content); with the following lines:
if(!empty($content['body']){
print render($content['body']);
} else {
print render($content['field_image']);
}
For more details: https://www.drupal.org/node/1323842
Hope this help!
As title said, how to set one header region different on each pages with Drupal 7.
For example:
On the page n1 : Header n1 (picture set by DP in Theme configuration)
On an other page : Header Different (idealy setted in a content type or in a tpl file)
If you know a module wich allow me to do this feature, thanks !
Regards
no module needed fore that. can't you just set a block to be displayed on one page only and another block on another page, both blocks you should place in header region.
this way its more consistent and you can more easily replace/adjust it
I created a View Argument and saved it as a block. I wanted to theme the Row style output and hence created a "views-view-fields--product-content-box2--block-1.tpl.php" and input my div tags with my class names and then included print commands to print the node body and node title. I did this based on the useful video # http://mustardseedmedia.com/podcast/episode23
My problem is that when the page containing the above view block is shown, the newly created "views-view-fields--product-content-box2--block-1.tpl.php" isnt being used. Instead the standard block.tpl.php is used.
I have emptied the cache to no avail.
Any ideas on how to get this working will be much appreciated.
The block.tpl.php file is used to theme the block as a whole, not the content within it.
To theme each row with a single tpl file, make sure your theme has a node.tpl.php file, set the view's row style to "node", and name your tpl node-view-product_content_box2.tpl.php.
I want to add some functionality to my Drupal 6 module: I want to insert an image next to certain node titles (it involves a span and an img tag). I thought I could do this by just implementing mymodule_preprocess_node() and modifying the title. However, Drupal strips out all tags to avoid XSS attacks.
I've seen many solutions that involve the theme layer (most commonly http://drupal.org/node/28537), but I want to do this in my module, not in the theme. I don't want to modify any .tpl.php files or template.php. Can anyone give me tips on how to do this?
You mention that you've tried preprocess_node(), and are correct that, if you are storing the img tag as part of the node title, Drupal will indeed strip that out, as it runs $node->title through check_plain in template_preprocess_node().
My suggestion would be to store the actual image as an image field (using some combination of the imagefield module and imagecache for sizing), set the display of that field to be hidden on the CCK display tab for the given content type, and then attach the image to be part of the $title variable in your module's preprocess function. This solution would also allow you to display that image next to the title in any views you may need to create.
By 'certain node titles' - do you mean all nodes titles from certain node types?
If so, you can likely style the node using only CSS. By default all nodes will have a class that corresponds to the node type. Using CSS background images, you can add an image to the node title.
Your module can call drupal_add_css and add in any required CSS into the page. This way, it is theme independent.
I think the easier way is with javascript / Jquery.
You create a Jquery script which is called only in certain types of nodes and pass the number of views from drupal to the jscript.
You can call drupal_add_js() inside your module_preprocess_node and pass a variable which contains the number of views or the image link to the script. Something like this:
<?php
drupal_add_js("var mymodule_imagelink = " . drupal_to_js($imagelink) . ";", 'inline');
drupal_add_js("my_js_file.js");
?>
then in my_js_file.js just change the text. There are several ways to acomplish this. For instance, you can do a search in the document and change the title to something else or you can use a determined ID, etc...
Find text string using jQuery?
http://api.jquery.com/replaceWith/
I tried to display body content with:
<?php print $node->content['body']['#value']; ?>
However, it doesn't display all body content, it just display first paragraph of body content, sometimes 2 paragraph if it is short :/
I need to print all body. how can I do that?
Thanks a lot! Appreciate helps!
The shortened body content hints on it being filled/rendered for 'teaser' view instead of 'full'. In what context do you issue this print statement?
EDIT: The node templates are usually used for both, teaser and full output, but the decision on what to use, as well as the population of the content entries in the node object happen outside of the node template files. Within the node template file, the variable $teaser will be TRUE, if the node is to be shown as a teaser.
So you need to check in what context your node template gets called, as you'll have to configure that context to render the node as 'full'. This could be in many places, depending on who is responsible to provide the nodes you want to theme, e.g. if the node template gets called from a view, you'll need to configure the view to use 'full page' output, if it comes from a module, you'll need to check with the module settings, etc...
In node.tpl.php try
<?php print $content ?>
However,
<?php print $node->content['body']['#value']; ?>
works for me as well.
To get control over your teaser length the master value is set with Post Settings.
(Length of trimmed posts)
To control this by node type try: http://drupal.org/project/teaserbytype
NOTE: Teasers are cached so you'll need to http://drupal.org/project/retease
However, if you want to just get it done in the node template you could run a node_load() and have everything... but that's not the best practice.
FYI: you can control what CCK fields show up in $content under Display Options.
PS: In teaser mode I often make use of truncate_utf8().