I have the following function:
public function make_order($id = null){
if($this->request->is('post')){
if(isset($id)){
$single_product = $this->Product->find('first', array('Product.id' => $id));
$this->placeOrder($single_product);
}else{
$product_array = $_SESSION['basket'];
foreach($product_array as $product){
$this->placeOrder($product);
}
}
}
}
private function placeOrder($product){
$order_array = array();
$order_array['Order']['Product_id'] = $product['Product']['id'];
$order_array['Order']['lejer_id'] = $this->userid;
$order_array['Order']['udlejer_id'] = $product['users_id'];
$this->Order->add($order_array);
}
Now these two function are not "connected" to a view but i still need to call them from within another view
For this ive tried the following:
<?php echo $this->Html->link(__('Bestil'), array('action' => 'make_order')); ?>
However this throws an error saying it couldnt find the view matching make_order and for good reason ( i havnt created one and i do not intend to create one)
My question is how do i call and execute this function from within my view?
At the end of your make_order function, you'll either need to:
a) specify a view file to render, or
b) redirect to a different controller and / or action, that does have a view file to render.
a) would look like this:
$this->render('some_other_view_file');
b) might look like this (note: setting the flash message is optional)
$this->Session->setFlash(__('Your order was placed'));
$this->redirect(array('controller' => 'some_controller', 'action' => 'some_action'));
You can turn auto-rendering off by setting $this->autoRender = false; in your controller's action (make_order() in this case). This way you don't need a view file, and you can output whatever you need.
The problem is that nothing will be rendered on the screen. Therefore, my advice is to have your "link" simply call a controller::action via AJAX. If that's not possible in your situation, then you'll have to either render a view in your make_order() method, or redirect to an action that will render a view.
Related
I'm brand new to CodeIgniter, so apologies if I'm missing something obvious here.
I'm comfortable with sending data from a controller to a view file using return view('default/blog/index', $data);. My issue is accessing the same data in a layout file which is extended by the view file using <?= $this->extend('layouts/default'); ?>.
For example, if I insert <?= $data['content'] ?> in my view file, it displays as expected. If I insert the same code in the layout file that is extended by my view file, I get the "Trying to access array offset on value of type null" exception.
What am I missing that will allow me to access my data from within the layout file?
Thanks in advance.
Update:
So in my BlogController I've got
class BlogController extends BaseController
{
public function index()
{
$model = new Blog();
$data = $model->getBlog();
return view('default/blog/index', ['data' => $data]);
}
public function item($slug = null){
$model = new Blog();
$data = $model->getBlog($slug);
return view('default/blog/item', ['data' => $data]);
}
}
And then in my item.php and index.php files, I have
<?= $this->extend('layouts/default', ['data' => $data]); ?>
My Blog Model's getBlog() method:
public function getBlog($slug = false)
{
if ($slug === false) {
return $this->orderBy('bs_created_dt', 'desc')->findAll();
}
return $this->where(['bs_slug' => $slug])->first();
}
When I use the debug toolbar to inspect the data, it is showing as expected, and I can display it in the view files, but not in the layout file.
In Codeigniter, you need to pass data also in an extended file called layouts.
Because you want to access data inside the extended file and for that, you just need to pass data to that file.
So replace the line of code of extended view with this :
$this->extend('layouts/default', ['data' => $data]);
Figured this out - absolute rookie mistake.
I'm working off of a pre-existing template, and the previous developer had overwritten the $data variable in layout file before where I was trying to use it.
I'm off to stand in the corner for a while.
In my web file, I have a route that accepts a $id as a value to be passed to a function within my PagesController. However, I want the function to still execute and show the intended form even when the $id is not passed to the function.
web.php file
Route::get('/request/{id}', 'PagesController#makeRequest');
PagesController.php file
public function makeRequest($id)
{
if(!empty($id)){
$target = Partner::find($id);
}
return view('pages.makeRequest')->with('target', $target);
}
makeRequest.blade.php
<input type="text" class="form-control" value="{{$target->inst_name}}" required disabled>
I want the page to display details from the database with the $id when provided or have empty spaces when the $id isn't provided.
As the Laravel Documentation states: Use Optional Parameters like this:
Route::get('/request/{id?}', 'PagesController#makeRequest'); //Optional parameter
Controller
public function makeRequest($id = null)
{
if(!empty($id)){
$target = User::find($id);
return view('pages.makeRequest')->with('target', $target);
} else {
return view('pageslist'); ///set default list..
}
}
This is the way I did it:
Route::get('index', 'SeasonController#index');
// controller
public function index(Request $request )
{
$id= $request->query('id');
}
The way you call it:
localhost/api/index?id=7
All your solutions were helpful. The main thing was that when I called just the view without passing $target to the view, the page displayed an error. So this is what I did.
Route::get('/request/{id?}', 'PagesController#makeRequest');
Then in the controller,
public function makeRequest(Request $request, $id=null)
{
if ($id != null) {
$target = Partner::find($id);
return view('pages.makeRequest')->with('target', $target);
}
return view('pages.makeNullRequest');
}
If you didn't understand what happened, I created a new view which had this instead of what I had posted in the question.
<input type="text" class="form-control" value="" required readonly>
Sorry I didn't update you guys in time. I think Jignesh Joisar came closest to helping me solve this issue. really appreciate all you guys. You're just awesome
You can use optional parameter :
Route::get('/request/{id?}', 'PagesController#makeRequest');
Now, as the parameter is optional, while defining the controller function you need to assign its default value to null in argument declaration.
<?php
public function makeRequest($id = null)
{
if($id){
$target = Partner::findOrFail($id);
return view('pages.makeRequest')->with(compact('target'));
}
// Return different view when id is not present
// Maybe all targets if you want
$targets = Partner::select('column1', 'column2')->get();
return view('pages.all')->with('targets');
}
I am using findOrFail instead of find. Its Laravel's very handy function which automatically throws a ModelNotFound exception and for frontend user throws a simple 404 page.
So if anyone is accessing www.url.com/request/2, its a valid id then it will show a valid page with data. If the accessed url is www.url.com/request/blahblah then it will throw 404. It avoids efforts of handling this manually.
For optional parameter pass id with ? in route and give $id = null in your function's parameter like this:
Route::get('/request/{id?}', 'PagesController#makeRequest'); //Optional parameter
makeRequest($id = null) {
// Code here...
...
}
in your routes file (web.php , as mentioned in your question)
Route::get('/request/{id?}', 'PagesController#makeRequest');
and in your controller PagesController.php
public function makeRequest($id = null)
{
}
To read more about this, just read https://laravel.com/docs/5.7/routing#parameters-optional-parameters
For me the answer was in the order that I listed the Routes in the routes file.
The routes file will call the first one that matches the pattern.
Route::get('/ohmy/{id?}', 'OhMyController#show');
Route::get('/ohmy/all', 'OhMyController#all'); //never gets called
Instead, put optional parameters at end of list:
Route::get('/ohmy/all', 'OhMyController#all');
Route::get('/ohmy/{id?}', 'OhMyController#show');
the answer has been said. just a side note: optional parameters won't work if you are using resource routes.
for example:
Route::resource('items',itemController::class)->except([
'create',
]);
Route::get('/items/create/{category_id?}',function($category_id = 'abc'){
dd($category_id);
});
if i go to " items/create/1 ", the result will be "1".
if i go to " items/create ", it will return 404. ( but we expect it to say "abc".)
this happens because other routes that start with "items" are expected to be generated from "resource" functionality.
so if you use resource routes, you should consider that.
It seems like i am not following the MVC design structure.
I'm making an ajax call from my view to a Controller function
Controller
public function actionGetClient()
{
$user = Client::model()->findByAttributes(array('email'=>$_POST['email'], 'password'=>$_POST['pass']));
echo $user->fullname;
}
View (the calling ajax)
CHtml::ajaxLink(
$text = 'get user',
$url = Yii::app()->createUrl('[my controller]/getClient'),
$ajax=array (
'type'=>'POST',
'data' => array('email'=>email, 'pass'=>pass),
'beforeSend' => "function( request )
{
$(\".result\").html(\"fetching...\")
}",
'success'=>"function(data){
$(\".result\").html(\"user is :\"+data)
}
"
));
Is it good to "echo" the $user->fullname inside the controller for the ajax success function to display it? My boss doesn't like it when i print stuff in my controller, how can i approach this
because when i use return instead, the ajax success gets a null value
return $user->fullname;
No,
It's not a good practice.
You need to create a view to use echo.
You can use return $this->renderPartial('VIEW_NAME'); to render a view without Layout.
You should write 'return' instead of 'echo'. 'echo' is not a good practice for ajax response. You don't need to make a new view for just return a name in your case.
public function actionGetClient()
{
$user = Client::model()->findByAttributes(array('email'=>$_POST['email'],'password'=>$_POST['pass']));
return $user->fullname;
}
No. A controller’s supposed to pass its results to a view for rendering.
I would avoid echoing in the controller what we usually do is have a ajax view folder and a json view and render with that so:
public function actionGetClient()
{
$user = Client::model()->findByAttributes(array(
'email'=>$_POST['email'],
'password'=>$_POST['pass']
));
$this->render("json",array("outputData"=>$user));
}
then add this to the controller as well:
public function getViewPath(){
if(Yii::app()->request->isAjaxRequest){
if(($module=$this->getModule())===null)
$module=Yii::app();
return $module->getViewPath().DIRECTORY_SEPARATOR."ajax";
}
return parent::getViewPath();
}
and in the ajax views folder add a json.php file like so
header('Content-Type: application/json');
// output data
echo json_encode($outputData);
please degug the code as I wrote it free hand. You can also set a marker in the controller like $viewPath and set it before the rendering
I'm trying to figure out how to use one of my view elements inside of a controller...
I know, I know: "Don't do that!" (99% of the time this is the correct answer)
But I think I actually have a good reason. The action is handling an AJAX request which returns markup. The returned markup is a list which I display everywhere else using an element. So in an effort to keep my code DRY, I think it's appropriate to do this here.
Is this possible?
Easy:
$view = new View($this, false);
$content = $view->element('my-element', $params);
Also:
DON'T DO THAT ANYMORE!!!
Sometimes, you need to render a CakePhp element from a view and inject its content into the page using AJAX the same time. In this case rendering element as a regular view from controller is better than creating a dedicated view that just contains <?php echo $this->element('some_element') ?>, and may be done this way:
<?php
public function ajax_action() {
// set data used in the element
$this->set('data', array('a'=>123, 'b'=>456, 'd'=>678));
// disable layout template
$this->layout = 'ajax';
// render!
$this->render('/Elements/some_element');
}
I know this is an old question and other people have already given basically the same answer, but I want to point out that this approach (provided by Serge S.) ...
<?php
public function ajax_action() {
// set data used in the element
$this->set('data', array('a'=>123, 'b'=>456, 'd'=>678));
// disable layout template
$this->layout = 'ajax';
// render!
$this->render('/Elements/some_element');
}
...is not a hacky workaround, but is in fact the recommended approach from the CakePHP docs for this common and legitimate use case:
If $view starts with ‘/’, it is assumed to be a view or element file
relative to the /app/View folder. This allows direct rendering of
elements, very useful in AJAX calls.
(Again: Credit to Serge S. for the code above)
$this->view = '/Elements/myelement';
You should use a client-side template. You should never return mark-up from a web service or API, just data. Have your JavaScript take the data, and then format it how you wish.
For example:
function getItems() {
$.get('/some/url', function(response) {
if (response.data.length > 0) {
for (var i = 0; i < response.data.length; i++) {
var item = response.data[i];
$('.results').append('<li>' + item.title + '</li>');
}
}
});
};
This is just an example written off the cuff. Obviously you’ll need to write your own implementation.
The way I did any ajax handling in Cake was to have my own AjaxController. Any interaction of ajax-kind goes there, which in-turn uses their own views (and view partials / elements). That way you can keep your code DRY and isolate and propagate all ajax use-cases there.
Example excerpt:
<?php
class AjaxController extends AppController {
/**
* (non-PHPdoc)
* Everything going to this controller should be accessed by Ajax. End of story.
* #see Controller::beforeFilter()
*/
public function beforeFilter() {
parent::beforeFilter();
$this->autoRender = false;
$this->layout = false;
if (!$this->request->is('ajax')) {
$this->redirect('/');
}
}
public function preview() {
if ($this->request->is('ajax')) {
$this->set('data', $this->data);
$this->render('/Elements/ajaxpreview');
}
}
?>
Here's the source: https://github.com/Sobient/dosspirit/blob/master/app/Controller/AjaxController.php
I setup the parseExtension for json in my cakephp 2.3.0. No errors are display. It works?
How can I test is?
In RoR ist ist very easy to test via
https://mydomain.de/view/4.json
How does it run on cakephp?
My View-Action is this.
public function view($id = null) {
if (!$this->Atla->exists($id)) {
throw new NotFoundException(__('Invalid atla'));
}
$options = array('conditions' => array('Atla.' . $this->Atla->primaryKey => $id));
$this->set('atla', $this->Atla->find('first', $options));
$this->Atla->id = $id;
$result = $this->Atla->read();
$this->response->type('json');
$this->response->body(json_encode($result));
return $this->response;
$this->set(compact('atlas'));
}
Why i always get an json-request?
If you use the _serialize key cakephp can automatically create json and xml views for you. I typically use the following to create json or xml views:
public function view() {
// data I want to display
$record1 = $this->ModelName->find('first', ...);
$this->set('record1', $record1);
$record2 = $this->ModelName->find('first', ...);
$this->set('record2', $record2);
// option 1: serialize the hard way
$this->set('_serialize', array('record1', 'record2'));
// option 2: serialize the easy way
$this->set('_serialize', array_keys($this->viewVars));
}
PS: the code after your return statement will never be executed.
you'll have create the view
app/View/Atlas/json/view.ctp
which is the view that is being used for .json requests.
Requests without .json will use the regular view file:
app/View/Atlas/view.ctp
There's more information on creating/using JSON and XML views here:
http://book.cakephp.org/2.0/en/views/json-and-xml-views.html#using-a-data-view-with-view-files
From that page the view.ctp may contain something like;
// View code - app/View/Posts/json/index.ctp
foreach ($posts as &$post) {
unset($post['Post']['generated_html']);
}
echo json_encode(compact('posts', 'comments'));
However, it really depends on what you're trying to achieve. If you will only use the 'Atlas/view' action for JSON responses, and won't be using HTML at all, sometimes you can get away with generating the response-body inside your Controller. Not very much 'in line' with MVC conventions, but it saves you from creating a view that does nothing more than echo json_encode($data); ;)
public function view($id)
{
$this->MyModel->id = $id;
$result = $this->MyModel->read();
$this->response->type('json');
$this->response->body(json_encode($result));
//Return reponse object to prevent controller from trying to render a view
return $this->response;
}
If you do want to use both 'HTML' and 'JSON', depending on the request (with/without .json extension), you should have two view files; 1 for JSON and 1 for HTML;
// This view will be used for your JSON requests
app/View/Atlas/json/view.ctp
// This view will be used for non-JSON (html) requests:
app/View/Atlas/view.ctp
In the json-view, output the data using json_encode(.....);
In the 'normal'/html view, just output normal data
In your controller, set the data as normal
public function view($id = null) {
$this->Atla->id = $id;
if (!$this->Atla->exists()) {
throw new NotFoundException(__('Invalid atla'));
}
$this->set('atla', $this->Atla->read());
}