I am looking for some advice on how I might fetch an array with a list of links to node types the currently logged in user is allowed to create.
My client wants these links to populate a custom dropdown list which sits on the user profile page.
Just in case I don't manage to talk him out of it, I would like some technique/information to go on.
You will have to create a custom module. If you are creating your own module, this short snippet will give you an array ($types) with the links to content types the logged in user can create (D6). If the user cannot create any content types it will show a message:
<?php
$types = array();
foreach (node_get_types('types', NULL, TRUE) as $type) {
if (node_access('create', $type->type)) {
$types[$type->type] = l($type->name, 'node/add/' . str_replace('_', '-', $type->type));
}
}
if (count($types) == 0) {
drupal_set_message('You cannot create any content types!', 'warning');
}
?>
Related
I have been working on a client's WordPress Website and last day my client want to hide navigation menu and pages from author/contributor categories.
I have searched and tried some of the plugin but didn't get the exact thing. Please let me know what should i use to hide some pages from user and from navigation.
Only Admin can see all the pages and other members should see only 1 section that is allowed to visible for them.
Thank You
use this plugin to manage All roll:
http://wordpress.org/plugins/user-role-editor/
Here is the Complete function for removing each Menu and submenu from wp-admin for another user:
function remove_menus() {
global $menu, $submenu;
$restricted = array(__('Dashboard'), __('Profile'), __('Users'), __('Tools'), __('Comments'), __('Settings'), __('Plugins')); //Here you can also define the name like Pages
end($menu);
while (prev($menu)) {
$value = explode(' ', $menu[key($menu)][0]);
if (in_array($value[0] != NULL ? $value[0] : "", $restricted)) {
unset($menu[key($menu)]);
}
}
unset($menu[5]); // this is just for example
unset($submenu['edit.php'][16]); // this is just for example
}
Now You have to put a conditon for other user i.e:
$thisusername = $current_user->user_login; // this is to get the current user login
if (($thisusername == "user123")) {
add_action('admin_menu', 'remove_menus');
}
Note: You can find many plugins but all of them are not in depth like this code.Well you can try this plugin to manage your user's roles.Capability Manager Plugin
I am trying to create a sampling process , where a user need to register or login to view my sample pdf documents , and once they register the user should be given a role as a sampler .
I am not a pro at wordpress but am learning . Can u guys suggest me some plugin or ideas on how to do it , i am using wordpress WordPress 3.3.1 .Thanks
http://codex.wordpress.org/Function_Reference/is_user_logged_in
http://codex.wordpress.org/Function_Reference/wp_login_form
As per your latest question, you will need to use these in tandem with each other, and make sure to define your redirect within your argument array to make sure the user is directed back to the page from which they logged in.
FINAL UPDATE: Enter this in your functions.php file and utilize as you see fit. This will allow you to output information according to the actual role of your user. This is also assuming that your plugin is creating actual custom roles. If not, you will have to utilize the Add Role Function:
function get_user_role()
{
global $current_user;
$user_roles = $current_user->roles;
$user_role = array_shift($user_roles);
return $user_role;
}
Then in the template of the page you want:
<?php
if(is_user_logged_in())
{
switch(get_user_role())
{
case 'customer' :
//CUSTOMER DISPLAY
break;
case 'subscriber' :
//SUBSCRIBER DISPLAY
break;
default :
//OTHER TYPE OF USER DISPLAY
}
}
else
{
//NO LOGGED USER DISPLAY
}
?>
This is a follow up question to Drupal Views exposed filter of Author name. The following question was answered and works. I can filter a view by user name. The user name is entered is entered by typing in a box and the box then auto completes. Rather then doing this I would like the list of users as a drop down. I only need one user to be selected. Do you know if this is possible?
You'll need a custom module for that.
I've done this for Drupal 7 this way: create a module, say, views_more_filters, so you have a views_more_filters.info file like this:
name = Views More Filters
description = Additional filters for Views.
core = 7.x
files[] = views_more_filters_handler_filter_author_select.inc
files[] = views_more_filters.views.inc
(file views_more_filters_handler_filter_author_select.inc will contain our filter handler).
A basic views_more_filters.module file:
<?php
/**
* Implements of hook_views_api().
*/
function views_more_filters_views_api() {
return array('api' => 3);
}
Then define your filter in views_more_filters.views.inc:
<?php
/**
* Implements of hook_views_data().
*/
function views_more_filters_views_data() {
return array(
'node' => array(
'author_select' => array(
'group' => t('Content'),
'title' => t('Author UID (select list)'),
'help' => t('Filter by author, choosing from dropdown list.'),
'filter' => array('handler' => 'views_more_filters_handler_filter_author_select'),
'real field' => 'uid',
)
)
);
}
Note that we set author_select as a machine name of the filter, defined filter handler ('handler' => 'views_more_filters_handler_filter_author_select') and a field we will filter by ('real field' => 'uid').
Now we need to implement our filter handler. As our filter functions just like default views_handler_filter_in_operator, we simply extend its class in views_more_filters_handler_filter_author_select.inc file:
<?php
/**
* My custom filter handler
*/
class views_more_filters_handler_filter_author_select extends views_handler_filter_in_operator {
/**
* Override parent get_value_options() function.
*
* #return
* Return the stored values in $this->value_options if someone expects it.
*/
function get_value_options() {
$users_list = entity_load('user');
foreach ($users_list as $user) {
$users[$user->uid] = $user->name;
}
// We don't need Guest user here, so remove it.
unset($users[0]);
// Sort by username.
natsort($users);
$this->value_options = $users;
return $users;
}
}
We haven't had to do much here: just populate options array with a list of our users, the rest is handled by parent class.
For further info see:
Views API
Where can I learn about how to create a custom exposed filter for Views 3 and D7? on Drupal Answers
Demystifying Views API - A developer's guide to integrating with Views
Sophisticated Views filters, part2 - writing custom filter handler (in Russian, link to Google translator)
Tutorial: Creating Custom Filters in Views
Yes, this is possible. Its not particularly tough to do this... but its slightly tedious. You need to create two views
The first view is a list of users on your system (a View of type Users). This user list is displayed as a dropdown instead of a list (using jump menu view style). Clicking on any user within this dropdown will call the second view with the uid (user id) of the selected user as the argument in the URL. This view is a block.
The second view is a simple Node listing. It is a page view at a particular URL. It takes 1 argument which is the uid (user id) of the user.
Detailed Steps
Download the Ctools module
http://drupal.org/project/ctools
Enable the Chaos Tools Module. This
module provides a Views Style Plugin
called "Jump Menu"
Create a new view of type Users and NOT type Node which you usually
create. In the fields add User:
Name and User: uid. For the
settings of User: uid, make sure
you click on Rewrite the output of
the field. The rewritten output of
the field should be
my_node_list/[uid]. Make sure you
select the exclude from display checkbox.
In the settings for Style in the view, select the Jump Menu style. Click on the settings for the style. Make sure the Path dropdown has User: uid choosen
Add a block display to the view. Name the block User Drop Down
Save the view
Add the block User Drop Down to any region in your theme e.g. Content Top (usually the best) or left sidebar. Make sure the block is only visible at the urls my_node_list/* and my_node_list by setting the block visibility settings
Now create another view of type Node. Add an argument field User: uid. Add the fields you are interested in e.g. Node: title, User: Name etc.
Add a page display. Let the page be at the url my_node_list
Save the view. Test the dropdown with its list of users on the system at http://yoursitename/my_node_list
http://drupal.org/project/better_exposed_filters
Check this one
I think you just have to choose "Taxonomy:term The taxonomy term ID" instead of "name".
Found a simple solution here. http://bryanbraun.com/2013/08/06/drupal-tutorials-exposed-filters-with-views
Using Drupal 6.x I have created two content types: Person and Event. Event has a custom field called Attendees (of type: Node Reference; unlimited number of values to person). When viewing a specific person how does one show all their events?
I have created a view (Personal Events) and added a block display. I enabled the block to show for content type Person. How should the view be defined? Or is there a better way?
Modules installed: CCK; Node Relationships; Views
I think one of these modules might be of help to you:
Reverse Node Reference
NodeReferrer
I have an answer to my own question. However, there maybe better answers... I can only hope.
Created content block (Personal Events)
Added this code to the body of the block. This code passes the node id argument to a view
<?php
if ( arg(0) == 'node' && is_numeric(arg(1)) && ! arg(2) ) {
$node = node_load(arg(1));
$args = array($node->nid );
$view = views_get_view('PersonalEvents');
print $view->preview('default', $args);
}
?>
Added this code to the Pages of the block [by selecting: Show if the following PHP code returns TRUE (PHP-mode, experts only)]... this drives the block to only appear person content.
<?php
//Read URL
$path=$_GET['q'];
//If URL is node page
if ( strpos($path,'node')===0){
//Parse URL to get nid
$links=explode("/",$_GET['q']);
$nid=$links[1];
//Load node
$node=node_load($nid);
//Display block only if node is of certain content type
if($node->type=='person'){
return TRUE;
}
}
return FALSE;
?>
Then created view with:
Style: Table
Relationship Content: Attendees (field_attendees); requires this relationship (checked); and Delta set to ALL.
Argument: Node: Nid; Relationship: Attendees; Hide view / Page not found (404) [selected]
Fields... simply selected Node Title and Date (for now)
Filter: Node Type = Event
Anyone have a better way?
I have a Drupal 6.14 site with Views module. I have a view and on the primary links I put a link to the view.
There is a way to hide the link in the primary menu only if the view is empty?
You could probably do this either via a theme or module implementation of preprocess_page (THEMENAME_preprocess_page(&$vars) or MODULENAME_preprocess_page(&$vars)), but mac above is correct in that views are not known to be empty or not until they are run, so there will be a performance hit.
Within the function, you should have access to the structured primary links array, so you can run the view:
$view = views_get_view('view_name');
// Swap out 'default' for a different display as needed. Also, $args are arguments, and can be left out if not applicable.
$output = $view->preview('default', $args);
if (empty($view->result)) {
// The view has no results, alter the primary links here to remove the link in question.
}
I am ready to be contradicted any moment as I never implemented anything like that, however I am under the impression that since views are essentially queries against the DB, you can't actually know if a view is empty until you actually invoke it.
Consider that - given you are speaking about primary links (shown on nearly every page of your site) this might be a serious performance hit, depending on the complexity of the view and on its "cacheability".
You should also consider whether the content of that view can be changed by other users browsing the site at the same time that "our" user: should the view become populated after "our" user has loaded the page, "our" user won't ever know.
As on how to achieve what you want, please see the accepted answer.
HTH!
I override views_embed_view() to only provide output if there is content, and then call my override from the theme layer:
function mymodule_embed_view($name, $display_id = 'default') {
// handle any add'l args (this hook supports optional params)
$args = func_get_args();
array_shift($args);
if (count($args)) {
array_shift($args);
}
$view = views_get_view($name);
$output = $view->preview($display,$args);
if ($view->result) {
return $output;
}
}
Then in the template file:
<?php
$view = mymodule_embed_view('view_name');
if (strlen($view) > 0) {
print $view;
}
?>