Elgg is build on the MVC framework. My main agenda is to be able to save the value selected from the dropdown list, after which is to then display the chosen value the item listing. The following code is actually constructed in PHP that is following the Elgg framework closely.
What I have managed to do is to make use of the existing Elgg framework to display the dropdown list. In which, the dropdown list is created by the creation of a form in the following directory: mod/plugin/views/default/forms/plugin/form.php. I have hence made use of the existing Elgg framework (input/dropdown)to create my dropdown list as a form.
Secondly, I have managed to save the values chosen in the dropdown list and display the value in a success message. This is done in the action directory which will allow the values to be saved into the database when user clicks on 'save' button.
Code for saving and displaying value:
<?php
/**
* Elgg options uploader/submit action
*
* #package ElggFile
*/
// get the input variables
$list = get_input('OptionItems');
$container_guid = (int) get_input('container_guid', 0);
if ($container_guid == 0)
{
$container_guid = elgg_get_logged_in_user_guid();
}
$my_select_guid = (int) get_input (file_guid);
//create a new my_select object
$my_select = new ElggObject();
$my_select -> dropdown = $list;
$my_select ->container_guid = $container_guid;
//save to database and get id of the new my_blog
$my_select_guid = $my_select->save();
if($my_select_guid){
system_message("Your action post = " . $list);
//to add new muy_select object to river
add_to_river('river/object/file/create', 'create', elgg_get_logged_in_user_guid(), $list->guid);
}
else{
register_error("Your action post is not saved");
}
However, at this point, I am stuck in displaying the chosen value of the dropdown list as an extended view, within the view/default/object/file/
How am I able to do this?
It's all in Elgg documentation about the views that I linked for you before: http://learn.elgg.org/en/1.12/guides/views.html#extending-views
Related
Is there a way to create/update form poll fields based on a content type? For example, I've got a post type called Candidate and I'd like like to dynamically update a poll list when a new Candidate is added, or when one is removed.
The reason I'm looking for this is because I'm creating a voting mechanism for this client and they have requested that users see an image, the name, and brief Bio of who they are voting for. My idea is to tie in the names so that I can target a hidden Gravity Form poll on page so when the voter clicks, it updates the corresponding named checkbox.
I can of course add each Candidate one by one, and then add each candidate one by one in the form, but was hoping there was a way to do that in code. So far I haven't found much other than filters in the Gravity Forms Documentation.
To reiterate, my question here is not the frontend connection, rather how to dynamically update/add an option in a poll field in a form when content is created for a Candidate post type.
Any help would be appreciated.
I believe the solution you're after is documented here:
http://www.gravityhelp.com/documentation/gravity-forms/extending-gravity-forms/hooks/filters/gform_pre_render/
Probably a combination of solution examples #1 and #2 on that page will fit your needs? So -
Create a category of standard Wordpress pages (category name: "Candidates"), and the pages would contain the candidate information (bio, photo, etc).
In your functions.php file for your current theme, you'd add something like the following to pull out that category's worth of posts:
// modify form output for the public page
add_filter("gform_pre_render", "populate_checkbox");
// modify form in admin
add_filter("gform_admin_pre_render", "populate_checkbox");
function populate_dropdown($form) {
// some kind of logic / code to limit what form you're editing
if ($form["id"] != 1) { return $form; }
//Reading posts for "Business" category;
$posts = get_posts("category=" . get_cat_ID("Candidates"));
foreach ($form['fields'] as &$field) {
// assuming field #1, if this is a voting form that uses checkboxes
$target_field = 1;
if ($field['id'] != $target_field) { break; }
// init the counting var for how many checkboxes we'll be outputting
// there's some kind of error with checkboxes and multiples of 10?
$input_id = 1;
foreach ($posts as $post) {
//skipping index that are multiples of 10 (multiples of 10 create problems as the input IDs)
if($input_id % 10 == 0) { $input_id++; }
// here's where you format your field inputs as you need
$choices[] = array('text' => $post->post_title, 'value' => $post->post_title);
$inputs[] = array("label" => $post->post_title, "id" => "{$field_id}.{$input_id}");
// increment the number of checkboxes for ID handling
$input_id++;
}
$field['choices'] = $choices;
$field['inputs'] = $inputs;
}
// return the updated form
return $form;
}
I'm new to yii and I don't understand the extensions much
but I used this extension called jmultiselect2side because I'm trying to make a site where users could reserve stuff like apparatuses in the lab
Anyway, I need a code that would get the Selected Items and then display them in another page for viewing purposes
I haven't put anything in the controller but the name of my controller and model is Apparatus
Here is my view:
<?php
$model= Apparatus::model()->findByAttributes(array('ApparatusCode'=>'1'));
// complete user list to be shown at multiselect order by ApparatusCode
$Apparatus= Apparatus::model()->findAll(
array('order' => 'ApparatusCode'));
?>
<center>
<?php
$this- >widget('application.extensions.jmultiselect2side.Jmultiselect2side',array(
'model'=>$model,
'attribute'=>'ApparatusName', //selected items
'labelsx'=>'Available',
'labeldx'=>'Selected',
'moveOptions'=>false,
'autoSort'=>'true',
'search'=>'Search:',
'list'=>CHtml::listData( // available items
$Apparatus,
'ApparatusCode',
'ApparatusName'),
));
?>
please help as soon as possible :/
put all above elements in a form. Set action for the form. Submit the form then the action in which you are handling this submit request, you can write there
if(isset($_POST))
{
foreach($_POST['Apparatus']['ApparatusName'] as $name)
{
do what ever you want
}
}
$name will represent the each selected item
I ended up using a session array and storing data there so that I can reference it again later. I just added my post data from each form into this array and referenced it later in my else block. Thanks for the help!
I am using CodeIgniter for a school project. I have some experience with PHP but am relatively new to this framework. I am having trouble using two forms on one page.
I have one form that displays a dropdown of artists. After clicking the submit button for this form, it updates the second form (another dropdown) on the same page with the portfolios belonging to the artist selected in the first dropdown.
I am trying to echo the values from each form just for testing purposes right now, I will implement other features later. My issue is that after my second form is submitted, the post value for the first form is changed. If I echo the selected value of the first form before submitting the second form, it shows the value that was selected. If I echo the value of the first form after both forms have been submitted, it shows the first available value from that dropdown.
I need to be able to take the values from both of these forms and then use them later after both forms have been submitted. So I can't have the values changing right when I need to use them, obviously, any help would be appreciated. Thank you much.
Controller
public function formtest(){
//Making a call to the model to get an array of artists from the DB
$data = $this->depot_model->get_artists_list();
$this->form_validation->set_rules('artist', 'Artist');// Commenting this out for now, 'required');
$this->form_validation->set_rules('ports', 'Portfolios', 'required');
if ($this->form_validation->run() == FALSE)
{
//Building the artists dropdown form
$data['data'] = form_open('formtest', 'class="superform"')
. form_label('Artist<br/>', 'artist')
. form_dropdown('artist', $data)
. form_submit('mysubmit', 'Submit')
. form_close();
//Setting up a temp array of the selected artist's portfolios
$ports = $this->depot_model->get_portfolios(url_title($data[$this->input->post('artist')]));
//Culling out the names of the portfolios from my temp array
$newdata = array();
foreach($ports as $port){array_push($newdata, $port['name']);}
//Building the artist's portfolio dropdown
$newdata['data'] = form_open('formtest', 'class="superform"')
. form_label('Portfolios<br/>', 'ports')
. form_dropdown('ports', $newdata)
. form_submit('mysubmit', 'Submit')
. form_close();
//Send the information to my view
$this->load->view('formtest', $data);
$this->load->view('formtest', $newdata);
}
else
{
//This echos out the first available value from my dropdown rather than the one I selected.
echo $data[$this->input->post('artist')];
echo "success";
}
}
The forms are separate, so when the second gets submitted, there is in effect no value received from the first form, as it isn't included as a field in the second. So you can do that, include say a hidden field in the second form that has the value of the artist. eg:
$newdata['data'] = form_open(
'formtest',
'class="superform"',
array('artist' => $this->input->post('artist'))
);
I have been trying to do display a custom field I created in the manage fields section of user accounts for nodes in addition to the profile page. The problem I am having with this code is that it will display the first field it finds and display that for every user, not the field for that particular user.
And ideas? I believe it's finding the first value in the array and not the value for the particular user in the array.
Here is m setup so far:
Added this to my template.php of my theme:
function mythemename_preprocess_node(&$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
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;
?>