using a view composer for the first time in Laravel. i have a sidebar that is included in every step of a submission form process a user goes through. i want the sidebar to have link that apply to the proper submission (i.e. if this is submission number 5, the links in the sidebar should all go to the edit function for submission 5.
i have the following code in my web.php:
View::composer('layouts.planbuilder', function($view){
$plansubmissions = PlanSubmission::find(3);
$view->with('plansubmissions', $plansubmissions) ;
}) ;
i am able to access the $plansubmissions variable, but of course this only applies to submission 3, which i hard coded in. is it possible to get the logic from another controller? i can't just get the user id with Auth because a user can have many submissions
View Composers also have access to the variables that were passed to the view itself, so if you are passing the submission to your view from the controller like so
return view('submissions.show', compact('submission');
Then in your composer you can assign it to $plansubmissions
$plansubmissions = $view->getData()['submission'];
Related
Currently my users must get the visit form given by Route::get then fill it in to get back a result view given by Route::post. I need to create a shareable link such as /account/search/vrm/{vrm} where {vrm} is the VRM that is usually filled in on the form page. This VRM then needs to redirected to Route::post as post data. This needs to be done by my controller. How can I do this in my controller?
Routes:
// Shows form view
Route::get('/account/search', 'User\AccountController#getSearch')->name('account.search');
// Shows result view
Route::post('/account/search', 'User\AccountController#runSearch');
// Redirect to /account/search as POST
Route::get('/account/search/vrm/{vrm}', function($vrm) { ???????? });
POSTs cannot be redirected.
Your best bet is to have them land on a page that contains a form with <input type="hidden"> fields and some JavaScript that immediately re-submits it to the desired destination.
You can redirect to a controller action or call the controller directly, see the answer here:
In summary, setting the request method in the controller, or calling a controller's action.
Ps: I don't want to repeat the same thing.
For those who comes later:
If you are using blade templating engine for the views, you can add '#csrf' blade directive after the form starting tag to prevent this. This is done by laravel to prevent cross site reference attacks. By adding this directive, you can get around this.
return redirect()->route('YOUR_ROUTE',['PARAM'=>'VARIABLE'])
I have a working project on Codeigniter 3. Now I have to build a FAQ page and I had this doubt: do I need a Controller for every URL?
It is, the FAQ page is a static page, but CodeIgniter generally routes URLs to Controllers, like domain/controller/method. But it seems a waste to build a Controller to only load the View.
No, it's not right way to make controller for every page. Just make one function which shows page by fetching data from database.
First of all make a table named pages in your database then save page_content, page_name, permalink for your different pages.
Now suppose your default controller is home, make a function in it with name page as below.
function pages( $permalink )
{
// get page data based on page_name passed in URL
$this->db->where( array( 'permalink' => $permalink ) );
$data['page'] = $this->db->get( 'pages' )->result();
// load view and pass page object to view
$this->load->view( 'view_file', $data );
}
Now same function will show different page content based on permalink passed in URL.
For example if URL is www.example.com/index.php/home/pages/faq then content of faq page will be shown.
I think this is slightly different to the usual controller passing data to the view. I have a Project which has one DocumentOne. Within my app, the user creates a Project. This then redirects them to the show page for this project.
So with the project created, and the user on the show page for that project, I display the project ID. I then provide a select menu where the user can select a Document to display. So say I am in Project with the ID of 1, I then decide to show DocumentOne for this project. This displays a form with inputs for DocumentOne.
When the user fills in the form and submits, the data is saved to the database. The Project ID is the foreign key for DocumentOne. The following route is set up for DocumentOne
Route::resource('projects.documentOne', 'DocumentOneController');
Now I have data for DocumentOne which is linked to the Project with an ID of 1. However, if I now go back to the projects show page and then select Document One from the dropdown again, all I see is an empty form. This is obviously because the controller for this is
public function show(Project $project)
{
return view('projects.show', compact('project'));
}
So I am never passing it data for DocumentOne because theoretically it is not created when the Project is first shown. What I want to do is when the Document is selected in the Projects show page, is to have the form populated with whatever is in the database for that Document. If nothing is in the database, then the form will be empty. I have a DocumentOne Controller, but I dont know if I can link this to the Projects show page. I was thinking about doing something like this in the DocumentOne controller
public function show(DocumentOne $documentOne)
{
return view('projects.show', compact('documentOne'));
}
But not sure this will work. Hope I have not been too confusing and you understand what I am attempting, hoping someone can offer advice on how best to handle this situation.
Thanks
In my previous project, I also deal with such requirement, I thought so. Here my solution to solve such requirement.
Actual code calling from ajax.
Routes
get('setFlashData',function(Request $request){
$final_response = array();
$data_information = $request->except('_token');
$request->session()->flash('cmg_quick_create_data', $data_information);
if($request->session()->has('cmg_quick_create_data')){
$final_response['result']['success'] = true;
}
return response()->json($final_response);
});
But according to you requirement:
$data_information = $request->except('_token');
$request->session()->flash('cmg_quick_create_data', $data_information);
My basic functionality was, to share form data from Quick Create Section which is pop-up form to Full create form section, and whenever user click to "Go To Full Form" button from pop up, ajax call mentioned function which will set the flash data and than on destination side I only check weather its contain the flash data or not. and deal according to data.
#if (Session::has('cmg_quick_create_data')) {
{!! Form::model(Session::get('cmg_quick_create_data'),["class"=>"form-horizontal","data-parsley-validate"=>"data-parsley-validate",'role'=>'form','files'=>true]) !!}
#else
{!! Form::open(["class"=>"form-horizontal","data-parsley-validate"=>"data-parsley-validate",'role'=>'form','files'=>true]) !!}
#endif
I can understand this solution might be different from you requirement but hope full to figure out your solution. Look forward to hearing from you if still unclear from my side.
I have an extension which should give the users (logged in as an Admin in the magento backend) the ability to change some configs in the frontend area. I want to have a link in the frontend which loads the config area via ajax and gives the user the possibility to edit&save this config in the loaded div. I want to use the magento backend forms for this so i don't have to code the forms myself.
My current approach has the link on the pages and loads via ajax the correct backend page (e.g. System > Configuration > Design). For this approach I created a Controller which extends the Mage_Adminhtml_Controller_Action. This Controller get the params from the ajax request and uses an action (like the editAction of the class Mage_Adminhtml_System_ConfigController) to get the right config page in the backend.
My Problems are:
- showing only the correct Area (I just want the user to edit only the section "themes" under System > Configuration > Design) everything else should be not available... so how to remove all the information around this config section?
The form needs the JS-variable Form_Key. How to get the current Form_Key (in the frontend)?
After the ajax has loaded the content the form doesnt get initialized correctly. So if I'm trying to submit the form my firebug says "JS-Error: configForm is not defined". How to solve this form initialising ? Any ideas?
I really hope anybody here can give me a hint how to solve this problems to get the backend config work in the frontend.
This is untested, but it should be enough to get you on the right track:
Output only a specific block
In the frontend most blocks are instantiated via layout XML. In the adminhtml area this is different, so you need to work with PHP instantiation much more.
In your AJAX action I assume you are currently calling loadLayout() and renderLayout().
To only output a specific section use this instead:
public function yourAjaxAction()
{
// assuming the required config section is set in the AJAX request
$sectionCode = $this->getRequest()->getParam('section');
$sections = Mage::getSingleton('adminhtml/config')->getSections();
$blockName = (string)$sections->frontend_model;
if (empty($blockName)) {
$blockName = Mage_Adminhtml_Block_System_Config_Edit::DEFAULT_SECTION_BLOCK;
}
$block = $this->getLayout()->createBlock($blockName)->initForm();
// Set the AJAX response content
$this->getResponse()->setBody($block->toHtml());
}
The form key
The form key can be fetched via
Mage::getSingleton('core/session')->getFormKey()
It must be present in the form posted back to the server. You can use the following code to create a HTML hidden field with the formkey:
// If loadLayout() was called:
$formkeyHtml = Mage::app()->getLayout()->getBlock('formkey')->toHtml();
// If working without layout XML:
$formkeyHtml = Mage::app()->getLayout()->createBlock('core/template', 'formkey')
->setTemplate('formkey.phtml') // adminhtml theme formkey
//->setTemplate('core/formkey.phtml') // frontend theme formkey
->toHtml();
Add configForm JavaScript
The configForm variable is an JS varienForm object of the DOM element containing the config fields.
It is instantiated using:
// config_edit_form is the CSS id
configForm = new varienForm('config_edit_form');
The varienForm declaration is in the file js/varien/form.js.
There also is some additional javascript used by the system configuration. Magento always adds in these blocks to set up the system config JS environment:
Mage::app()->getLayout()->getBlock('js')->append(
$this->getLayout()->createBlock('adminhtml/template')
->setTemplate('system/shipping/ups.phtml')
);
Mage::app()->getLayout()->getBlock('js')->append(
$this->getLayout()->createBlock('adminhtml/template')
->setTemplate('system/config/js.phtml')
);
Mage::app()->getLayout()->getBlock('js')->append(
$this->getLayout()->createBlock('adminhtml/template')
->setTemplate('system/config/applicable_country.phtml')
);
I hope that gets you started.
I am trying to use CodeIgniter and jQuery-ui dialog to create a modal window with form to update user information.
The process should be like:
1. Press a button on a view page.
2. A modal window pops up.
3. Inside the window is a form that a user can fill.
4. If the user filled something before, the information should be shown in corresponding field
5. Click the update button on the modal window to save the changes to database.
Can anyone provide a good sample of this process?
I used ajax to pass the data but it didn't work when I was trying to update the data to the database. It would be nice if an example of how to pass data from ajax to php and how php handle that.
Thanks,
Milo
well the jquery bit for post(), get(), ajax() works the same in any measure you would normally use it.. key difference here is with CI you can't post directly to a file-name file-location due to how it handles the URI requests. That said your post URL would be the similar to how you would access a view file normally otherwise
ie: /viewName/functionName (how you've done it with controllers to view all along. post, get, ajax doesnt have to end in a extension. I wish I had a better example then this but I can't seem to find one at the moment..
url = '/home/specialFunction';
jQuery.get(url, function(data) {
jQuery("#div2display").html(data);
});
in the case of the above you notice despite it not being a great example that. you have the url with 2 parameters home and specialFunction
home in this case is the controller file for home in the control folder for the home file in views the specialFunction is a "public function" within the class that makes the home controller file. similar to that of index() but a separate function all together. Best way I have found to handle it is through .post() and a callback output expected in JSON cause you can form an array of data on the php side json_encode it and echo out that json_encode and then work with that like you would any JSON output. or if your just expecting a sinlge output and not multiples echoing it out is fine but enough of the end run output thats for you to decide with what your comfortable doing currently. Hopefully all around though this gives you some clairity and hopefully it works out for you.