custom drupal 6 is not showing on the page - php

I recently started working for this company and I am trying to modify the php code that was already created to display blocks on a page. I am using my localhost and wamp to test and develop my code. the first block that I created worked as the only thing that I did was to copy and modify the title of the code but now that I am trying to do the same for second and for some reason is not working. on my drupal 6 localhost block page I could see the block that I created and the region assign to it but is still not visible in the page. What am I missing?
Below is the code that I am using. Any help will be appreciated.
on the Page specific visibility settings this is the php code :
<p><?php return _admin_block_visibility('breeding_race_results'); ?></p>
$blocks['race_results'] = array('info' => t('Race Results'),
'status' => 1,
'region' => 'middle',
'visibility' => 2,
'weight' => -100,
'pages' => '<?php return _admin_block_visibility(\'race_results\'); ?>',
);
case 'race_results':
$block['subject'] = t('Race Results');
$block['content'] = admin_race_results_block_content();
break;

Generally, in Drupal, you don't need to write a code to have things displayed. Use Drupal core blocks management (Structure > Blocks) or use Context module to tell the system where you want certain elements like blocks to appear (what page or what type of pages etc.). Also, use Views module to create a dynamic lists of things to be displayed anywhere in the system.
Also, if a block is not showing where it suppose to, add there some text text and check if that will work. You will want to figure out first if the problem is with the code in the block (that makes the results invisible) or the problem is that the block is not rendering at all.

Related

How am I supposed to be using filter_xss? Even though I'm using it, coder gives me an issue

+269: [critical] Potential problem: drupal_set_message
http://api.drupal.org/api/function/drupal_set_message/() only accepts
filtered text, be sure all !placeholders for $variables in t
http://api.drupal.org/api/function/t/() are fully sanitized using
check_plain http://api.drupal.org/api/function/check_plain/(),
filter_xss http://api.drupal.org/api/function/filter_xss/() or
similar.
Which pertains to this code:
drupal_set_message(t('Batch complete! View/Download !results', array(
'!results' => filter_xss(l(t('simple results'), file_create_url($filename))),
)), 'info');
What's going wrong?
The method you're using is under the 'DO NOT DO THESE THINGS' portion of Dynamic or static links in translatable strings. You need to change it to one of the approved methods. For reference:
<?php
// DO NOT DO THESE THINGS
$BAD_EXTERNAL_LINK = t('Look at Drupal documentation at !handbook.', array('!handbook' => ''. t('the Drupal Handbooks') .''));
$ANOTHER_BAD_EXTERNAL_LINK = t('Look at Drupal documentation at the Drupal Handbooks.');
$BAD_INTERNAL_LINK = t('To get an overview of your administration options, go to !administer in the main menu.', array('!administer' => l(t('the Administer screen'), 'admin'));
// Do this instead.
$external_link = t('Look at Drupal documentation at the Drupal Handbooks.', array('#drupal-handbook' => 'http://drupal.org/handbooks'));
$internal_link = t('To get an overview of your administration options, go to the Administer screen in the main menu.', array('#administer-page' => url('admin')));
?>

Serve an HTML5 application as a Drupal Module

I manage a Drupal 7 based website for my College's Student's Association. Most of it is standard static pages. Each year we run a room ballot for people to choose their rooms and this process will need an application to display current room allocations (in real time) and wiki style information about what the different rooms are like.
I need to be able to serve up static pages of HTML, javascript and css; bypassing the theming module. I need the relative addressing in the html page which serves as the root of the application to work properly (e.g. "javascript/app.js" should pick up that file from within the module). I then need to serve up json data from php using all the drupal APIs for permissions and database access etc.
I have a fair bit of experience in HTML, Javascript etc. and some in PHP, but I'm fairly new to Drupal module development.
You should create a custom module as you suggest, and separately put your HTML5 application in a sub-folder of the module. When it's accessed it will use the same relative paths as you'd normally expect so javascript/app.js will work if the file exists in the path under your HTML5 app's folder.
For the JSON data your custom module will look something like this:
function mymodule_menu() {
$items['my/app/data'] = array(
'page callback' => 'mymodule_ajax_callback',
'access callback' => TRUE,
'type' => MENU_CALLBACK
);
return $items;
}
function mymodule_ajax_callback() {
$type = $_POST['type'];
$nodes = db_query("SELECT nid, title FROM {node} WHERE type = :type", array(':type' => $type))->fetchAllKeyed();
drupal_json_output($nodes);
drupal_exit();
}
That code defines a menu path (using hook_menu()) at mp/app/data which uses mymodule_ajax_callback() as it's page callback.
mymodule_ajax_callback() simply grabs all nodes from the database that match the type parameter passed in the AJAX $_POST and outputs their id and title in a JSON string to the page (which will then be returned as your AJAX response when you request /my/app/path).
Hope that helps

How to display a Drupal 6.20 menu in WordPress?

Is is possible to display (via php) the main menu of a Drupal 6.20 site in a WordPress theme template file located in a subdirectory on the same domain?
Right now, I'm displaying the menu by copying the static html from the Drupal site and adding it to header.php in the WordPress template in the site located in mydomain.com/blog/. But of course that's not going to work when another menu item is added to the Drupal site, or the Drupal menu is changed in any way.
So is there a Drupal php function that will pull the menu into the WP file?
Failing that, is there a way with php to parse a Drupal page for the html of the menu (yes, this would be ugly) and display it in WP?
The first part of the challenge is to output only the menu, with as little (or none) of the surrounding HTML as possible, so you have as little work to do in parsing the HTML as possible.
The second part is to take that output from Drupal and actually display it on your WordPress site.
You could add the Drupal database as a secondary database in WordPress using the a new instance of the $wpdb object, write the query to get the right content from the tables, and format the results. That could work, but might be overkill.
An alternative workable option may be to use JSON to format the output of the primary links, using the drupal_json function in Drupal, then consume the JSON feed in Wordpress.
I'm assuming:
you have admin access to login to the Drupal site, which you'll need to create nodes, and clear the theme cache
you want to output the Primary Links menu, which 90%+ of Drupal sites use. This is probably true, but it is possible your site uses custom menus. If so, this is still possible, you'd just write slightly different code in step 3.
The steps would be:
Create a Drupal node (you can call it anything, it's just a placeholder)
Get the node id of your dummy page (ie., node/234). From the node id, create a one-off page template in your Drupal site's themes folder. It should be called page-node-xxxx.tpl.php, with xxxx being your node id
Add this code to page-node-xxxx.tpl.php:
<?php
drupal_json(menu_navigation_links(variable_get('menu_primary_links_source', 'primary-links')));
?>
This will create a JSON feed of your menu items.
Clear the theme cache of your Drupal site by visiting http://yoursite.com/admin/build/themes and visit http://yoursite.com/node/xxxx to see the raw JSON feed.
You should now be able to use a jQuery method like $.getJSON or $.ajax in your Wordpress theme to consume and display the JSON feed, or possibly use json_decode and curl to output your array as HTML.
A good thing about Drupal's drupal_json function is that it already sends the correct JSON headers, so now all you have to do is write the jQuery or PHP that does what you need.
I'm assumed you are more of a Wordpress specialist and have a working knowledge of Drupal but maybe not a lot of familiarity with its inner workings. So, sorry if it seemed too basic (or not basic enough :).
The Drupal theming engine is very modular - you may be able to make an appropriate PHP call into Drupal to get just the menu rendered, then emit that HTML as a part of your WordPress page.
g_thom's answer is very good and if you wish to create a very simple module to output the main navigation you can write something like this:
<?php
function getmenus_help($path, $arg) {
// implementing the help hook ... well, not doing anything with it just now actually
}
function getmenus_all() {
$page_content = '';
$page_content = json_encode(menu_navigation_links(variable_get('menu_primary_links_source', 'primary-links')));
// fill $page_content with the menu html
print $page_content;
return NULL;
}
function getmenus_menu() {
$items = array();
$items['getmenus'] = array(
'title' => 'Get Menus',
'page callback' => 'getmenus_all',
'access arguments' => array('access getmenus'),
'type' => MENU_CALLBACK,
);
return $items;
}
// permissions
function getmenus_perm() {
return array('access getmenus');
}
In your PHP code you can then write something like:
function primary_links() {
$primary_links = file_get_contents(SITE_URL . '/getmenus');
$primary_links = json_decode($primary_links);
$primary_links = (array)$primary_links;
$i = 0;
$last = count($primary_links);
$output = '';
foreach ($primary_links as $pm) {
$href = $pm->href;
if (strpos($pm->href, 'http://') === FALSE) {
if ($pm->href == '<front>') {
$href = SITE_URL . '/';
} else {
$href = SITE_URL . '/' . $pm->href;
}
}
$output .= '
<li>
'.$pm->title.'</li>';
$i++;
}
return $output;
}
I hope this helps!
PS: Make sure you update the module's permissions to allow anonymous users to have access to the path you set in your module - otherwise you will get a 403 Permission Denied.

SugarCRM - How to set the default homepage for all users

I am using SugarCRM Version 5.2.0k (Build 5837). I would like to be able to set a default home page (with dashlets I've created myself) that will be the same for all users, can anyone advice on best way to do this?
Thanks in advance for your help
I'd like to know how to do this too... see here for some ideas, but it's clear that it's not a supported feature.
I wonder if you can write a module that installs a hook for post user creation (assuming that this hook is provided) and then populate the appropriate part of the user preferences table when the hook is invoked. Of course, your module will probably break with each upgrade of SurgarCRM, so this might be more trouble than it i worth.
Edit:
I had a look at the Dash Manager module that is referenced in the thread I linked to above. It's approach is to copy the preferences of the admin user to all other users when the administrator clicks a link in the admin page. So, the admin user is used as a sort of template for other users. Rudimentary solution, but not a bad start - using a template user and treating the preferences (as stored in the DB table) as opaque seems like the way to go.
It's quite easy to do it.
I have done it in SugarCRM 6.5.23.
Here I have mentioned steps to do it:
Just copy sugarcrm_root/modules/Home/index.php and paste it in SugarCRM_root/custom/modules/Home/index.php.
Now you can customize it's behavior as you want.
You can remove default dashlets and add your own dashlets by creating one file at SugarCRM_root/custom/modules/Home/dashlets.php and add this code in it:
<?php
unset($defaultDashlets);
$defaultDashlets = array(
'CustomDashlet' => 'ModuleName',
'UpcomingAppointmentsDashlet' => 'Meetings', //Example
);
Once you do this thing still you have 3 dashlets left in your hook code you can remove it if it's needed code for that hook is like this:
$dashlets[create_guid()] = array(
'className' => 'iFrameDashlet',
'module' => 'Home',
'forceColumn' => 0,
'fileLocation' => $dashletsFiles['iFrameDashlet']['file'],
'options' => array('titleLabel' => 'LBL_DASHLET_DISCOVER_SUGAR_PRO',
'url' => '...',
'height' => 315,
));
Hope this will help you. :)

Drupal - Automate a Content Form Submission

I would like to programatically (using php) fill out an existing drupal form to create a content type that is included in a contributed module.
Details: The module is SimpleFeed and the content type is Feed. I would like to call the module's functions to accomplish this. The method I am interested in is hook_insert which appears to require vid and nid which I am unsure what these are.
Any help is appreciated.
can you provide a bit more information (which modules?). generally, i'd probably suggest calling the modules functions to create the content type, instead of trying to pass it through a form programatically. this way you don't have to worry about implementation, and can trust that if the module works, it'll work for your script too :)
of course this does tie your module to theirs, so any changes in their functions could affect yours. (but then again, you run that risk if they update their database structure too)
ex.
// your file.php
function mymodule_do_stuff() {
cck_create_field('something'); // as an example, i doubt this
// is a real CCK function :)
}
edit: vid and nid are node ID's, vid is the revision id, and nid is the primary key of a particular node. because this is an actual node, you may have to do two operations.
programatically create a node
you'll have to reference the database for all the exact fields (tables node and node_revisions), but this should get you a basic working node:
$node = (object) array(
'nid' => '', // empty nid will force a new node to be created
'vid' => '',
'type' => 'simplefeed'. // or whatever this node is actually called
'title' => 'title of node',
'uid' => 1, // your user id
'status' => 1, // make it active
'body' => 'actual content',
'format' => 1,
// these next 3 fields are the simplefeed ones
'url' => 'simplefeed url',
'expires' => 'whatever value',
'refresh' => 'ditto',
);
node_save($node);
now i think it should automatically call simplefeed's hook_insert() at this point. if not, then go on to 2. but i'd check to see if it worked out already.
call it yourself!
simplefeed_insert($node);
edit2: drupal_execute() isn't a bad idea either, as you can get back some validation, but this way you don't have to deal with the forms API if you're not comfortable with it. i'm pretty sure node_save() invokes all hooks anyhow, so you should really only have to do step 1 under this method.
The drupal api provides drupal_execute() to do exactly this. I would suggest you avoid calling the functions directly to create the node (unless there is a performance reason). By using drupal_execute() all the proper hooks in other modules will be called and your code is far more likely to continue to work through future versions of drupal.
Note that a classic bug in using this method is not first calling something like
module_load_include('inc', 'node', 'node.pages')
which will load the code for your node creation form.
Calling node_save directly is generally considered deprecated and could leave you with broken code in future versions of drupal.
There is a nice example at this lullabot post

Categories