I have a custom page information.php with a link in every course in Moodle, where is printed the summary of every course.
To show the course name I have in information.php
global $DB, $COURSE;
$course = $COURSE->fullname;
The problem is, that it prints only the information from the first course id:
Coursename1
and not according to the current course, where I clicked on the link.
What I am doing wrong?
Thanks in advance.
It would be better to pass the course id to your information page.
In the original page display the link:
$url = new moodle_url('\local\yourplugin\information.php', array('id' => $course->id));
echo html_writer::link($url, get_string('informationlink', 'local_yourplugin'));
Then in your information.php page:
$courseid = required_param('id', PARAM_INT);
// Use get_course() for performance.
$course = get_course($courseid);
// Use require_login($course) to test if the user can see this course.
require_login($course);
Related
I currently have a form in laravel on whos submission the following methods run:
public function validateSave() {
$qualitycheck = new QualityCheck();
$qualitycheck['site-name'] = Request::input('site-name');
$qualitycheck['favicon'] = Request::has('favicon');
$qualitycheck['title'] = Request::has('title');
$qualitycheck['image-optimization'] = Request::has('image-optimization');
$qualitycheck->save();
Session::flash('quality-data', $qualitycheck);
return redirect('/');
}
So i have the below line that passes the data to the next page:
Session::flash('quality-data', $qualitycheck);
But what i would really want to do is, when the form is submitted, i would really just want to show a link on the next page , which will be coded like so:
#if(Session::has('quality-data'))
Submited Quality Check
#endif
Now on click on the link , i would like to show a view with all the data that the user submitted in the form , How do i do this ? I.E. How do i pass the data form from the <a> to the view that will show up when clicked on the <a> ??
So just to put things into perspective, this is how it works now:
STEP-1 :: User submits form , data is flashed to next page.
STEP-2 :: Data user submits is shown on this page.
How i want it to work is:
STEP-1 :: User submits form , data is flashed to next page.
STEP-2 :: A link is shown to the user(Only if user clicks on the link we move to the next step).
STEP-3 :: Data user submited in first step is shown on this page.
Next time you code, please follow the below coding practices.
Prefer using create() function of model.
Put all your request data that is to be used in one variable (like $input)
Prefer using route names like route('route.name') instead of strings inside redirection() function.
Please replace your function with
public function validateSave() {
$inputs = [
'site-name' => request()->get('site_name'),
'favicon' => request()->has('facvicon'),
'title' => request()->has('title'),
'image-optimization' => request()->has('image-optimization')
]);
$qualityCheck = QualityCheck::create($inputs);
$flashMessage = '<a href=' . route('quality.check.show', $qualityCheck) . '>Submitted Quality Check</a>'
Session::flash('quality-data', $flashMessage);
return redirect(route('home.index'));
}
And ensure you have something like this in your routes file.
Route::get('quality-check/{id}', 'QualityCheckController')->name('quality.check.show');
Let me know if something doesn't work...
try this one.
<a href="{{url('routename')}}">
How to get joomla content using ajax? (I want to show content of specyfic page in popup), this is my code: (called by ajax)
$option = JRequest::getCmd('option');
$view = JRequest::getCmd('view');
if ($option=="com_content" && $view=="article") {
$ids = explode(':',JRequest::getString('id'));
$article_id = $ids[0];
$article =& JTable::getInstance("content");
$article->load($article_id);
echo '<h2>'.$article->get("title").'</h2>';
echo $article->get("introtext"); // and/or fulltext
}
This works fine only for artilces, but the problem is when for example I want to show category, or component
Please see your if condition it checks if option is equal to com_content & view is equal to article only. If view contains category it won't work. So add the conditions in if statement so that your code gets executed.
for category you need to add view=category & like for other components as well.
I have been trying to do display a custom field I created in the manage fields section of user accounts for nodes in addition to the profile page. The problem I am having with this code is that it will display the first field it finds and display that for every user, not the field for that particular user.
And ideas? I believe it's finding the first value in the array and not the value for the particular user in the array.
Here is m setup so far:
Added this to my template.php of my theme:
function mythemename_preprocess_node(&$vars) {
global $user;
$user = user_load($user->uid); // Make sure the user object is fully loaded
$team = field_get_items('user', $user, 'field_team');
if ($team) {
$vars['field_team'] = $team[0]['value'];
}
}
Then, added this to my node.tpl.php in order to display it on nodes.
if (isset($field_team) && !empty($field_team)) :
echo '$field_team.'</div>';
endif;
UPDATE:
Found my own aswer here:
http://drupal.org/node/1194506
Code used:
<?php
$node_author = user_load($node->uid);
print ($node_author->roles[3]);
print ($node_author->field_biography['und'][0]['value']);
?>
You can use drupal's 'Author Pane' module for that. Try this:
http://drupal.org/project/author_pane
i have url like this :
http://quickstart.local/public/category1/product2
and in url (category1/product2) numbers are id , categorys and products fetched from database attention to the id
id is unique
i need to the sensitive url like zend framework url. for example :http://stackoverflow.com/questions/621380/seo-url-structure
how i can convert that url to the new url like this
is there any way?!!
You'll need to store a unique value in your database with a field name such as 'url' or something similar. Every time you generate a new product you will have to create this unique url and store it with the product information. A common way to do this is to take the name of the product and make it url friendly:
public function generateUrl($name)
{
$alias = str_replace(' ', '-', strtolower(trim($name)));
return preg_replace('/[^A-Za-z0-9-]/', '', $alias);
}
Calling this method:
$url = $this->generateUrl("My amazing product!");
echo $url;
will output:
my-amazing-product
You'll need to check that the output from this function does not already exist in the database as you will use this value to query on instead of the id.
If you apply this logic to the categories as well, you can have easily readable and descriptive urls like the one below. You may need to tweak your routing before this works correctly though.
http://quickstart.local/public/awesome-stuff/my-amazing-product
You could use ZF's Zend_Controller_Router_Route. For example, to make similar url to those used by SO, one could define a custom route in an application.ini as follows (assuming you have controller and action called questions and show respectively):
resources.router.routes.questions.route = '/questions/:id/:title'
resources.router.routes.questions.type = "Zend_Controller_Router_Route"
resources.router.routes.questions.defaults.module = default
resources.router.routes.questions.defaults.controller = questions
resources.router.routes.questions.defaults.action = show
resources.router.routes.questions.defaults.id =
resources.router.routes.questions.defaults.title =
resources.router.routes.questions.reqs.id = "\d+"
Having such a route, in your views you could generate an url as follows:
<?php echo $this->url(array('id'=>621380,'title' => 'seo url structure'),'questions');
// results in: /myapp/public/questions/621380/seo+url+structure
//OR if you really want to have dashes in your title:
<?php echo $this->url(array('id'=>621380,'title' => preg_replace('/\s+/','-','seo url structure'),'questions');
// results in: /myapp/public/questions/621380/seo-url-structure
Note that /myapp/public/ is in the url generated because I don't have virtual hosts setup on my localhost nor any modifications of .htaccess made. Also note that you don't need to have unique :title, because your real id is in :id variable.
As a side note, if you wanted to make it slightly more user friendly, it would be better to have your url as /question/621380/see-url-structure rather than /questions/621380/see-url-structure. This is because under this url you would have only one question, not many questions. This could be simply done by changing the route to the following resources.router.routes.questions.route = '/question/:id/:title'.
EDIT:
And what to do with categories and products that you have in your question? So, I would define a custom route, but this time using Zend_Controller_Router_Route_Regex:
resources.router.routes.questions.route = '/questions/(\d+)-(d+)/(\w*)'
resources.router.routes.questions.type = "Zend_Controller_Router_Route_Regex"
resources.router.routes.questions.defaults.module = default
resources.router.routes.questions.defaults.controller = questions
resources.router.routes.questions.defaults.action = show
resources.router.routes.questions.map.1 = category
resources.router.routes.questions.map.2 = product
resources.router.routes.questions.map.3 = title
resources.router.routes.questions.reverse = "questions/%d-%d/%s"
The url for this route would be then generated:
<?php echo $this->url(array('category' => 6213,'product' => 80,'title' => preg_replace('/\s+/', '-', 'seo url structure')),'questions' ); ?>
// results in: /myapp/public/questions/6213-80/seo-url-structure
Hope this will help or at least point you in the right direction.
Friends a newbie question.........I need help in getting the URL of a specific Menu itemID. The situation is like this:
I am running Joomla and asking for a user to input for a menu ID and choose a layout for that menu ID.
I want to do something else with this URL of the Menu itemID.
How can I get the URL of this Menu itemID provided by the user?
For Example if the user input is liek $this->get ('menulayoutid'>; and he inputs and ID of 54 then how do I get the URL for Menu ID 54.
Please note: I want to get this URL from within my PHP file and not in the browser so that I can use the value of that URL for some other purpose.
Kindly help.
$itemid = JRequest::getVar('Itemid');
$application = JFactory::getApplication();
$menu = $application->getMenu();
$item = $menu->getItem($itemid);
$link = new JURI($item->link);
$link->setVar('ItemId', $itemid);
Source: http://forum.joomla.org/viewtopic.php?p=1836005
However, we get the Itemid from anywhere (user input, from our own developed module using the "menu item" field type in the xml file as described in the Joomla Docs - Standard form field types)
// get the menuItemId from wherever...
// as described above or as in other posts here and do whatever with that!
$menuItemId = 'fromWherever'; // as an example "107";
// build the link to the menuItemId is just easy and simple
$url = JRoute::_('index.php?Itemid=' . $menuItemId);
i think if we need only a link to a specific menu id, this is the best solution, because we have absolutely less requests and a clean code
this works also in Joomla 3.0, 3.1
I just want to add that if you need to target a specific menu you pass the menu name as an argument to getMenu().
$itemid = JRequest::getVar('Itemid');
$application = JFactory::getApplication();
$menu = $application->getMenu( 'menu-name' );
$item = $menu->getItem($itemid);
$link = new JURI($item->link);
$link->setVar('ItemId', $itemid);
I'm not sure if Joomla changed the way this works since 2.5 or even 1.7 but I spent the worse half of 2 hours looking for this.
Hopefully it helps someone.
$menuID = $params->get('menuItem'); // from module field menu ex. '105'
$js = new JSite;
$menu = $js->getMenu();
$link = $menu->getItem($menuID)->route;
//Returns URL Friendly Link -> menu/article
//Then format it ->
$link = 'http://www.yoursite.com/index.php/'.$link;
echo 'Borrowed Menu Link Path";
When you need to get your active menu item ID in Joomla to display some specific content for only that menu item or just to show the ID of the menu item, insert the following code where you wish to display the active menu item ID:
<?php
$currentMenuId = JSite::getMenu()->getActive()->id;
echo $currentMenuId;
?>