I am trying to merge the contents of 2 arrays then use usort to get the posts with most views.
Trying to use usort to sort the contents of an array.
Am getting the follow error:
("Notice: Undefined property: Acme\DemoBundle\Entity\Article::$getViews in /.../PageController.php line 15")
Can someone point out what I am doing wrong?
Sort function inside of controller
private static function popularSort($articles, $posts, $articles2, $posts2)
{
return $articles->getViews() == $posts->getViews() ? 0 : ( $articles->getViews() < $posts->getViews()) ? 1: -1;
}
Sidebar action
$articles = $this->getDoctrine()->getRepository('AcmeDemoBundle:Article')
->getArticles();
$posts = $this->getDoctrine()->getRepository('AcmeDemoBundle:Post')
->getPosts();
$popular = array_merge($articles, $posts);
usort($popular, array($this, 'popularSort'));
getViews is a getter method for accessing the property views of the Entities Post and Article. So while accessing it u should access it as $articles->getViews().
But if you simply want to compare the property views of the two entities compare them using their property name instead of their getter
Assuming views as the name of the property, the call should be something like:
$posts->views and $articles->views.
You don't have a property called getViews in your Article class.
You probably have a property called views and a method getViews meaning you should call the actual method with the brackets like $article->getViews().
Related
I can't find any info about using $this->Cookie in view cells.
When i wrote code like this, error will arise:
namespace App\View\Cell;
use Cake\View\Cell;
class CityCell extends Cell {
public function display() {
$this->Cookie->config('encryption', false);
$cookie = $this->Cookie->read('city');
}
}
and the error is:
Error: Call to a member function read() on null
So can we use cookie in view cells ?
Thank You.
That of course won't work, view cells do not support the use of components, altough they can be thought of like "mini-controllers", they are not actually controllers in the terms of CakePHPs MVC system.
Depending on whether the cookies are encrypted, you can either use the request object to fetch them in your cell
$this->request->cookie('cookieName')
or
$this->request->cookies
see also API > \Cake\Network\Request::cookie()
or, in case they are encrypted, you have grab them via the Cookie component, and then for example pass them down from your controller to the view, and finally into the cell like
controller
public function controllerAction() {
// ...
$this->set('cookie', $this->Cookie->read('cookieName'));
}
cell
public function display($cookie) {
// ...
}
view
$this->cell('CellName', ['cookie' => $cookie]);
see also Cookbook > Views > View Cells > Passing Arguments to a Cell
I have this error:
Notice (8): Undefined variable: informations [APP/View/Information/index.ctp, line 2]
I have this function in my controller
public function index($slug){
$this->layout = 'sbhealth';
$this->loadModel('Menu');
$informations = $this->Menu->findBySlug($slug);
if($informations){
$this->set('index', $informations);
}else{
return $informations = "not find";
}
}
I need to pass datas to my view "index.ctp" index.ctp will then find the ID in my array to pass it to an Element. but the View doesn't recognize "$informations"
This probably doesn't constitute something that should be on StackOverflow, since it's specified in great detail in the book, but...
// in your controller
$this->set('informations', $informations);
The first item is the name of the variable that will be available in the view.
The second item is the value you want to put in that variable.
When I use the Backbone.Collection.where function to filter the collection I get an array of models as return value but not an other filtered collection object. So I can't use other collection functions with that.
What is the purpose of such behavior?
where isn't the only method that returns an Array. where returns a new Array because you definitely don't want it mutating your existing Collection automatically. Also, many times you may want the result in Array form.
For whatever reason, the BB devs decided that it was better to return a new Array rather than a new Collection. One thought could be that, perhaps the returned data would be used in a different type of Collection. Another reason could be so that you always know what is returned from one of these methods. 2+ types of collections will ALWAYS return Arrays from these types of methods rather than having to try and inspect via instanceof or something else that isn't very reliable.
Edit
In addition, you COULD make your collections behave in a manner where you return new Collections. Create a base Collection to do something like this:
// Override the following methods
var override = ["where","find",...];
var collectionProto = Backbone.Collection.prototype;
BaseCollection = Backbone.Collection.extend({});
for (var key in collectionProto) {
if (collectionProto.hasOwnProperty(key) && override.indexOf(key) > -1) {
BaseCollection.prototype[key] = function () {
return new this.constructor(collectionProto[key].apply(this, arguments);
};
}
}
Instead over extending off Backbone.Collection, extend off BaseCollection.
Note that you can still use most of the underscore utilities on arrays. Here's how to use each() after a filter()
_.each( MyCollection.filter( filter_fn() {} ), each_fn() {} )
I searched a lot but I couldn't find on How to use the find('all') in Views as used in Rails, but here I'm getting the error "Undefined property: View::$Menu [APP\Lib\Cake\View\View.php, line 804]"
'Menu' is the model which I'm using to fetch data from the menus table.
I'm using the below code in views:
$this->set('test',$this->Menu->find('all'));
print_r($test);
Inside your Menu model create a method, something like getMenu(). In this method do your find() and get the results you want. Modify the results as you need and like to within the getMenu() method and return the data.
If you need that menu on every page in AppController::beforeFilter() or beforeRender() simply do
$this->set('menu', ClassRegistry::init('Menu')->getMenu());
If you do not need it everywhere you might go better with using requestAction getting the data using this method from the Menus controller that will call getMenu() from the model and return the data. Setting it where you need it would be still better, if you use requestAction you also want to cache it very likely.
TRY TO NOT RETRIEVE DATA WITHIN VIEW FILE. VIOLATION OF MVC RULE
try this in view file:
$menu = ClassRegistry::init('Menu');
pr($menu->find('all'));
In AppHelper ,
Make a below function
function getMenu()
{
App::import('Model', 'Menu');
$this->Menu= &new Menu();
$test = array();
$test = $this->Menu->find('all');
return $test;
}
Use above function in view like :
<?php
$menu = $html->getMenu();
print_r($menu);
?>
Cakephp not allow this .
First create the reference(object) of your model using ClassRegistry::init('Model');
And then call find function from using object
$obj = ClassRegistry::init('Menu');
$test = $obj->find('all');
echo ""; print_r($test); `
This will work.
If I have a person model with first_name and last_name, how do I create and display a full_name? I would like to display it at the top of my Edit and View views (i.e. "Edit Frank Luke") and other places. Simply dropping echoes to first_name and last_name isn't DRY.
I'm sorry if this is a very simple question, but nothing has yet worked.
Thank you,
Frank Luke
Edit for clarity: Okay, I have a function on the person model.
function full_name() {
return $this->Person->first_name . ' ' . $this->Person->last_name;
}
In the view, I call
echo $person['Person']['full_name']
This gives me a notice that I have an undefined index. What is the proper way to call the function from the view? Do I have to do it in the controller or elsewhere?
If what you are wanting is just to display a full name, and never need to do any database actions (comparisons, lookups), I think you should just concatenate your fields in the view.
This would be more aligned with the MVC design pattern. In your example you just want to view information in your database in a different way.
Since the action of concatenating is simple you probably don't save much code by placing it in a separate function. I think its easiest to do just in the view file.
If you want to do more fancy things ( ie Change the caps, return a link to the user ) I would recommend creating an element which you call with the Users data.
The arrays set by the save() method only return fields in the datbase, they do not call model functions. To properly use the function above (located in your model), you will need to add the following:
to the controller, in the $action method:
$this->set( 'fullname', $this->Person->full_name();
// must have $this-Person->id set, or redefine the method to include $person_id
in the view,
echo $fullname;
Basically, you need to use the controller to gather the data from the model, and assign it to the controller. It's the same process as you have before, where you assign the returned data from the find() call to the variable in the view, except youre getting the data from a different source.
There are multiple ways of doing this. One way is to use the afterFind-function in a model-class.
See: http://book.cakephp.org/view/681/afterFind.
BUT, this function does not handle nested data very well, instead, it doesn't handles it al all!
Therefore I use the afterfind-function in the app_model that walks through the resultset
function afterFind($results, $primary=false){
$name = isset($this->alias) ? $this->alias : $this->name;
// see if the model wants to attach attributes
if (method_exists($this, '_attachAttributes')){
// check if array is not multidimensional by checking the key 'id'
if (isset($results['id'])) {
$results = $this->_attachAttributes($results);
} else {
// we want each row to have an array of required attributes
for ($i = 0; $i < sizeof($results); $i++) {
// check if this is a model, or if it is an array of models
if (isset($results[$i][$name]) ){
// this is the model we want, see if it's a single or array
if (isset($results[$i][$name][0]) ){
// run on every model
for ($j = 0; $j < sizeof($results[$i][$name]); $j++) {
$results[$i][$name][$j] = $this->_attachAttributes($results[$i][$name][$j]);
}
} else {
$results[$i][$name] = $this->_attachAttributes($results[$i][$name]);
}
} else {
if (isset($results[$i]['id'])) {
$results[$i] = $this->_attachAttributes($results[$i]);
}
}
}
}
}
return $results;
}
And then I add a _attachAttributes-function in the model-class, for e.g. in your Person.php
function _attachAttributes($data) {
if (isset($data['first_name']) && isset($data['last_name'])) {
$data['full_name'] = sprintf("%s %s %s", $data['first_name'], $data['last_name']);
}
return $data;
}
This method can handle nested modelData, for e.g. Person hasMany Posts then this method can also attachAttributes inside the Post-model.
This method also keeps in mind that the linked models with other names than the className are fixed, because of the use of the alias and not only the name (which is the className).
You must use afterFind callback for it.
You would probably need to take the two fields that are returned from your database and concatenate them into one string variable that can then be displayed.
http://old.nabble.com/Problems-with-CONCAT-function-td22640199.html
http://teknoid.wordpress.com/2008/09/29/dealing-with-calculated-fields-in-cakephps-find/
Read the first one to find out how to use the 'fields' key i.e. find( 'all', array( 'fields' => array( )) to pass a CONCAT to the CakePHP query builder.
The second link shows you how to merge the numeric indexes that get returned when you use custom fields back into the appropriate location in the returned results.
This should of course be placed in a model function and called from there.