I have a Problem with the ChildrenOf(pagename) and multilanguage.
When i use
<% control ChildrenOf(aktuelles/messen).Limit(2) %>
It works in German but not English.
So i added
public function PageByLang($url, $lang) {
$SQL_url = Convert::raw2sql($url);
$SQL_lang = Convert::raw2sql($lang);
$page = Translatable::get_one_by_lang('SiteTree', $SQL_lang, "URLSegment = '$SQL_url'");
if ($page->Locale != Translatable::get_current_locale()) {
$page = $page->getTranslation(Translatable::get_current_locale());
}
return $page;
}
to my Page Controller and in the Template:
<% control ChildrenOf(PageByLang(aktuelles/messen, de_DE)).Limit(2) %>
But its still not working.
I dont know if its a good approach but i solved my problem with overriding the ChildrenOf in my Page Controller:
public function ChildrenOf($parentRef) {
$parent = SiteTree::get_by_link($parentRef);
if(!$parent && is_numeric($parentRef)) {
$parent = DataObject::get_by_id('SiteTree', $parentRef);
}
$locale = Translatable::get_current_locale();
if($parent && $parent->getTranslation($locale))
return $parent->getTranslation($locale)->Children();
}
Related
I'm trying to understand Joomla language and I have this situation:
In a models/calcoloonline.php I have this function
public function estraivariabili()
{
$db = JFactory::getDBO();
// Put the result into a variable first, then return it.
$value = $db->setQuery("SELECT * FROM #__calcolo_imposte")->loadObjectList();
if ($value != NULL)
{
return $value;
}
else
{
return JFactory::getApplication()->enqueueMessage(JText::_('COM_CALCOLO_IMPOSTE_IMPORTI_NON_DEFINITI'), 'type');
}
}
This works perfectly but I'd like that after check if the return is NULL I want to hide display default.php and show only the message on JText.
How can I do this?
For your purpose, you just return the $value from the model function and call the function at view.html.php's display() function.
At default.php file check the availability of the $value and show your contents.
For example, you store the data at view.php.html. It looks like
public function display($tpl = null)
{
$model = $this->getModel();
$this->value = $model->estraivariabili();
return parent::display($tpl);
}
And your default.php file would be
<?php if (!empty($this->value)) { ?>
<h1>The value is not empty.</h1>
<?php } else {
// value not found :(
JFactory::getApplication()->enqueueMessage(JText::_('NOT_FOUND_MESSAGE'), 'warning');
} ?>
I'm trying to pass a variable from a controller to another controller I tried using
Redirect::to('dashboard/'.$ssid.'/')->with(compact('wname'))
but does not work any idea how can I achieve this?
here is my code
Route
Route::get('dashboard/{ssid}/', 'HomeController#showDash');
LoginController
public function post_index()
{
if(Auth::attempt($credentials)){
$users = User::where('username','=',$email)->get();
foreach ($users as $value):
$activated = $value['a_status'];
$wname = $value['wholename'];
endforeach;
if($activated == 1):
$red= Redirect::to('dashboard/'.$ssid.'/')->with(compact('wholename'));
else:
$red= View::make('login');
endif;
return $red;
}
}
HomeController
public function showDash($ssid,$wholename)
{
foreach ($wholename as $userVal):
$fn = $userVal['firstname'];
$ln = $userVal['lastname'];
endforeach;
return View::make('dashboard')->with(compact('fn'));
}
The error I'm having is that Missing argument 2 for HomeController::showDash() as per laravel's debugger..
Updated answer based on the comments:
public function post_index()
{
if(Auth::attempt($credentials)){
$users = User::where('username','=',$email)->get();
foreach ($users as $value){
$activated = $value['a_status'];
$wname = $value['wholename'];
}
if($activated == 1) {
Redirect::to('dashboard/'.$ssid.'/')->with(['wholename' => $wholename]);
}
return View::make('login');
}
public function showDash($ssid)
{
$wholename = (Session::has('wholename')) ? Session::get('wholename') : [];
foreach ($wholename as $userVal) {
$fn = $userVal['firstname'];
$ln = $userVal['lastname'];
}
return View::make('dashboard')->with(compact('fn'));
}
Everything else can stay as is.
Update: fixed erroneous space in 'whole name' (autocorrect did that, sorry).
I am writing a custom component. I have the view employees. Under this view, I have two layouts, default and modal.
I have a menu item in the toplevel of the main menu, Employees that points to my employee view:
index.php?option=com_mycomponent&view=employees which resolves to domain.com/joomla/employees and displayes the default view as expected.
Now, inside my component I want to link to the modal view, and I do so using JRoute and this url:
index.php?option=com_mycomponent&view=employees&layout=modal
Which resolves to
domain.com/joomla/employees/modal
And produces this error:
500 - View not found [name, type, prefix]: modal, html,
mycomponentView
If I visit index.php using index.php?option=com_mycomponent&view=employees&layout=modal my modal view is displayed.
I have also found that visiting domain.com/joomla/employees/employees/modal displays the correct layout. It is as if joomla is forgetting what view is associated with the menu item at /joomla/employees, and instead looks for the view "modal" unless the extra "employees" is provided in the url.
Also worth noting, domain.com/joomla/employee?layout=modal works fine as well.
Here is what I have for my router.php. This was file was generated for me using the component generator at j-cook.pro.
<?php
defined('_JEXEC') or die;
function MycomponentBuildRoute(&$query){
$segments = array();
if(isset($query['view']))
{
$view = $query['view'];
$segments[] = $view;
unset( $query['view'] );
}
if(isset($query['layout']))
{
$segments[] = $query['layout'];
unset( $query['layout'] );
}
if(isset($query['id']))
{
if(in_array($view, array('edit','view','view','editfacility','view','edit','client','editclient','viewposition','editposition','edit','view','edit','view','view','edit','view','edit','view','edit','view','edit')))
{
$segments[] = (is_array($query['id'])?implode(',', $query['id']):$query['id']);
unset( $query['id'] );
}
};
return $segments;
}
function MycomponentParseRoute($segments)
{
$vars = array();
$vars['view'] = $segments[0];
$nextPos = 1;
if (isset($segments[$nextPos]))
{
$vars['layout'] = $segments[$nextPos];
$nextPos++;
}
if(in_array($vars['view'], array('edit','view','view','editfacility','view','edit','client','editclient','viewposition','editposition','edit','view','edit','view','view','edit','view','edit','view','edit','view','edit'))
&& isset($segments[$nextPos]))
{
$slug = $segments[$nextPos];
$id = explode( ':', $slug );
$vars['id'] = (int) $id[0];
$nextPos++;
}
return $vars;
}
So it is hard to provide an exact answer for this without knowing all the different ways that you want to have urls be parsed. But I will try to give a hint at what will solve this present situation (without hopefully introducing too many new issues!)
The basic issue is that the BuildRoute side does not get a view value so it is not included in the url. On the one hand it is not necessary, because it is in the menu. But it makes it a little harder to parse, so option one is to force there to be a view if you can by changing the top function to start like this:
function MycomponentBuildRoute(&$query){
$segments = array();
if(isset($query['view']))
{
$view = $query['view'];
$segments[] = $view;
unset( $query['view'] );
}
else
{
$app = JFactory::getApplication();
$menu = $app->getMenu();
$active = $menu->getActive();
if ($view = $active->query['view']) {
$segments[] = $view;
}
}
...
In this way, if there is a menu item for this and it has a view we will tack it on. This should generate domain.com/joomla/employees/employees/modal.
You could also probably do this logic on the parse side too. This would go instead of the other option above:
function MycomponentParseRoute($segments)
{
$vars = array();
$app = JFactory::getApplication();
$menu = $app->getMenu();
$active = $menu->getActive();
if ($active->query['view']) {
$vars['layout'] = $segments[0];
$nextPos = 1;
} else {
$vars['view'] = $segments[0];
$nextPos = 1;
if (isset($segments[$nextPos]))
{
$vars['layout'] = $segments[$nextPos];
$nextPos++;
}
}
... continue with final check for id
I would probably use the second option but both are an option. By the way, you are also likely to run into issues if you try to use an id without setting a layout.
This is sample of function in the Staff controller for this question
function newStaff()
{
$data = array();
$data['departmentList'] = $this->department_model->list_department();
$data['branchList'] = $this->branch_model->list_branch();
$data['companyList'] = $this->company_model->list_company();
$this->load->view('staff/newstaff', $data);
}
function add_newStaff()
{
//when user submit the form, it will call this function
//if form validation false
if ($this->validation->run() == FALSE)
{
$data = array();
$data['departmentList'] = $this->department_model->list_department();
$data['branchList'] = $this->branch_model->list_branch();
$data['companyList'] = $this->company_model->list_company();
$this->load->view('staff/newstaff', $data);
}
else
{
//submit data into DB
}
}
From the function add_newStaff(), i need to load back all the data from database if the form validation return false. This can be troublesome since I need to maintain two copy of codes. Any tips that I can use to prevent this?
Thanks.
Whats preventing you from doing the following
function newStaff()
{
$data = $this->_getData();
$this->load->view('staff/newstaff', $data);
}
function add_newStaff()
{
//when user submit the form, it will call this function
//if form validation false
if ($this->validation->run() == FALSE)
{
$data = $this->_getData();
$this->load->view('staff/newstaff', $data);
}
else
{
//submit data into DB
}
}
private function _getData()
{
$data = array();
$data['departmentList'] = $this->department_model->list_department();
$data['branchList'] = $this->branch_model->list_branch();
$data['companyList'] = $this->company_model->list_company();
return $data;
}
Alternately you change the action your form submits to so that it points to the same service you use for the initial form request with something like the following. This would also mean that you'd have the POST values retained between page-loads if you wanted to retain any of the submitted values in your form.
function newStaff()
{
// validation rules
if ($this->validation->run() == TRUE)
{
//submit data into DB
}
else
{
$data = array();
$data['departmentList'] = $this->department_model->list_department();
$data['branchList'] = $this->branch_model->list_branch();
$data['companyList'] = $this->company_model->list_company();
$this->load->view('staff/newstaff', $data);
}
}
I want to be able to user query strings in this fashion. Domain.com/controller/function?param=5&otherparam=10
In my config file I have
$config['base_url'] = 'http://localhost:8888/test-sites/domain.com/public_html';
$config['index_page'] = '';
$config['uri_protocol'] = 'PATH_INFO';
$config['enable_query_strings'] = TRUE;
The problem that I am getting is that form_open is automatically adding a question mark (?) to my url.
So if I say:
echo form_open('account/login');
it spits out: http://localhost:8888/test-sites/domain.com/public_html/?account/login
Notice the question mark it added right before "account".
How can I fix this?
Any help would be greatly appreciated!
The source of your problem is in the Core Config.php file where the CI_Config class resides. The method site_url() is used by the form helper when you are trying to use form_open function.
Solution would be to override this class with your own. If you are using CI < 2.0 then create your extended class in application/libraries/MY_Config.php, otherwise if CI >= 2.0 then your extended class goes to application/core/MY_Config.php.
Then you need to redefine the method site_url().
class MY_Config extends CI_Config
{
function __construct()
{
parent::CI_Config();
}
public function site_url($uri='')
{
//Copy the method from the parent class here:
if ($uri == '')
{
if ($this->item('base_url') == '')
{
return $this->item('index_page');
}
else
{
return $this->slash_item('base_url').$this->item('index_page');
}
}
if ($this->item('enable_query_strings') == FALSE)
{
//This is when query strings are disabled
}
else
{
if (is_array($uri))
{
$i = 0;
$str = '';
foreach ($uri as $key => $val)
{
$prefix = ($i == 0) ? '' : '&';
$str .= $prefix.$key.'='.$val;
$i++;
}
$uri = $str;
}
if ($this->item('base_url') == '')
{
//You need to remove the "?" from here if your $config['base_url']==''
//return $this->item('index_page').'?'.$uri;
return $this->item('index_page').$uri;
}
else
{
//Or remove it here if your $config['base_url'] != ''
//return $this->slash_item('base_url').$this->item('index_page').'?'.$uri;
return $this->slash_item('base_url').$this->item('index_page).$uri;
}
}
}
}
I hope this helps and I think you are using CI 2.0 that wasn't officially released, this was removed in the official CI 2.0 version
Simpler might be to just set follwing in your config.php
$config['enable_query_strings'] = FALSE;
Was the solution in my case.
If you want to use query string in your url structure, then you should manually type your url structure in the following order:
<domain.com>?c={controller}&m={function}¶m1={val}¶m2={val}
in the action of the resepective controller you should get the parameter as $_GET['param1']
your code now should look like this
form_open(c=account&m=login¶m1=val)
Please let me know if it doesnt work for you.