In my project I have this method in my view:
public function elegirSeleccionados(){
$this->assignRef('pagination', $this->get('pagination'));
$this->assignRef('items', $this->get('recientes'));
$this->assignRef('list', $this->get('list'));
parent::display();
}
assignRef is deprecated / removed in this Joomla 3.
What should I use instead?
assignRef() and assign() are not needed anymore, since Joomla 1.6+
requires at least PHP 5.2 (PHP5 uses assign by reference).
Use in view.html.php
$this->pagination = $this->get('pagination')
and in the template just call $this->pagination.
To get your skills up to date check the official Joomla! Documentation
txs for this, but in my view (default.tpl) zeh val will not been shown
view.html.php
$this->my_string = "ledl";
<br>
return parent::display($tpl);
default.tpl
<bold>{$this->my_string}</bold>
Related
I'm facing strange case. I face an error in production env not while in dev it's working fine.
Development:
Laravel 5.4.28
PHP 7.0.13
MYSQL 5.7.17
Production:
Laravel 5.4.28
PHP 7.2.1
MYSQL 5.7.20
In implementation code. I used:
namespace App;
use Illuminate\Support\Facades\Storage;
use Laravel\Scout\Searchable;
use Illuminate\Database\Eloquent\Model;
class Artwork extends Model
{
use Searchable;
In development it works fine. But in production it gives me this error:
count(): Parameter must be an array or an object that implements Countable
in Builder.php (line 936)
as you can see in this pic:
Any idea what is the reason behind this? and how to fix?
Put this code at the beginning of your route file, it will work fine
if(version_compare(PHP_VERSION, '7.2.0', '>=')) {
error_reporting(E_ALL ^ E_NOTICE ^ E_WARNING);
}
This is a documented change in PHP 7.2. You need to either update Laravel to 5.6 or downgrade PHP to version 7.1.
Replace
$originalWhereCount = count($query->wheres);
by
$originalWhereCount = count((array)$query->wheres);
in
\vendor\laravel\framework\src\Illuminate\Database\Eloquent\Builder.php
I was facing similar issue in Laravel 5.6. Where I was getting error for object based array. I knew that data in that particular variable will always remain object so i used to convert the object to array. Here is code sample:
$objectData = (array)$objectData;
echo "Total Elements in array are: ".count($objectData);
My server was on PHP 7.1 when I updated to PHP 7.2 I got the same issue.
After searching I found why this occurs. (This occurs because of a PHP update.).
so in my case, the error is solved by typecasting.
I just update all code where I used to count
Before
//this is before
count($adminDetails)
After updated
//after update all i typecast all the code where i used count
count((array)$adminDetails)
Goodluck
This error occurs because you are using a higher PHP version and your Laravel application is on an older PHP version.
✅ Simple solution:
Open: app/Providers/AppServiceProvider.php
And in: public function register() { ... } function add following code:
if(version_compare(PHP_VERSION, '7.2.0', '>=')) {
error_reporting(E_ALL ^ E_NOTICE ^ E_WARNING);
}
In php 7.2+ count does not work on relation objects, you need to use:
$model->relation()->exists()
Not this (less than php 7.2):
count($model->relation)
i ran into the same problem (PHP 7.2 + Laravel 5.3) but i don't see any "good" answers here. For me, the problem occurs when i tried to start a Builder from a scope method on the model: SomeModel::forUser() calls scopeForUser(). Trying to build a new Query, it trips on a count($this->wheres) that gets no initial value (null). Because the magic static call to the scope starts the builder, no other conditions have been placed in the object so the property is still null at that point.
i thought it's worth sharing my solution first, then perspective on why i consider it better than Ben's answer. It's not personal, i just disagree.
Solution
i took a cue from this answer about overriding some of the core Illuminate\Database classes...
Extend Illuminate\Database\Eloquent\Model
Mine is App\Overrides\Database\Eloquent\Model
Extend Illuminate\Database\Eloquent\Builder
Mine is App\Overrides\Database\Eloquent\Builder
Extend Illuminate\Database\Query\Builder
Can you guess? App\Overrides\Database\Query\Builder
Tell Laravel to use YOUR Eloquent\Model:
config/app.php 'aliases' array, replace the 'Eloquent' value
with your Eloquent\Model FQN
My Model:
namespace App\Overrides\Database\Eloquent;
/*
* Notes:
* * Using replacement Query\Builder with ALIAS
* * Use of Builder in this class is MY Eloquent\Builder
*/
use App\Overrides\Database\Query\Builder as QueryBuilder;
use Illuminate\Database\Eloquent\Model as EloquentModel;
class Model extends EloquentModel
{
public function newEloquentBuilder($query)
{
return new Builder($query);
}
protected function newBaseQueryBuilder()
{
$conn = $this->getConnection();
$grammar = $conn->getQueryGrammar();
return new QueryBuilder($conn, $grammar, $conn->getPostProcessor());
}
}
My Eloquent\Builder:
namespace App\Overrides\Database\Eloquent;
use Illuminate\Database\Eloquent\Builder as EloquentBuilder;
class Builder extends EloquentBuilder
{
public function __construct($query)
{
parent::__construct($query);
/*
* FIX #1: Set properties treated AS arrays
* to empty arrays on construct.
*/
$this->wheres = [];
// Any other properties treated as arrays should also be initialized.
}
}
My Query\Builder:
namespace App\Overrides\Database\Query;
use Illuminate\Database\Query\Builder as QueryBuilder;
class Builder extends QueryBuilder
{
public function __construct()
{
parent::__construct(...func_get_args());
/*
* FIX #2: Set properties treated AS arrays
* to empty arrays on construct.
*/
$this->wheres = [];
// Any other properties treated as arrays should also be initialized.
}
}
This safely preserves the framework's functionality, since the only actual change you're making is initializing properties that should have been in the first place. Everything else will pass instanceof checks used for dynamic loading and dependency injection.
Opinion
While i agree with #ben-harold about every comment he made saying "NEVER edit vendor code," i disagree with the "solution." It's an oversimplification to a much more complex problem.
Upgrade Laravel: to ensure support for PHP 7.2, jumping up several minor versions - if not major releases - is impractical for a lot of teams. As a long term objective, yes of course. As something i can do to get rid of the bug for my deadline? Nope. Upgrading takes a lot of planning and frequently a lot of rewrites as structures, names, and functionality change. It's something to prioritize, but not a need-it-now answer.
Downgrade PHP: same problem. Downgrading into PHP 5.x means A) PHP is EOL, which may be a deal breaker for a lot of customers who have security policies, and B) any usage of PHP 7.x language features have to be scrapped. As with upgrading the framework this is very likely to cause a lot of headaches. It's also an even less useful solution, since walking backward in the language just puts you farther behind and will require more long-term effort.
place the below line ob code before the class name in your controllers
if (version_compare(PHP_VERSION, '7.2.0', '>=')) {
// Ignores notices and reports all other kinds... and warnings
error_reporting(E_ALL ^ E_NOTICE ^ E_WARNING);
// error_reporting(E_ALL ^ E_WARNING); // Maybe this is enough
}
I was facing the same issue with an external created table (Not using migration or command),
After creating the model, I just assigned a table name, but the problem was in my model protected $fillable where I assign string instead of array and error occurred.
There is 2 possible solution for that.
Assign an array to your protected $fillable = ['filed1', 'filed2'];
Remove protected $fillable completely (Not Recommended)
class Abc extends Model
{
protected $table = 'cities';
protected $fillable = ['field1','field2', ...];
}
Model looking for countable parameter:
class ClassName extend Model {
protected $fillable=['column_name']; // column in DB of Model is in array
}
Before
count($collection['colors'])
Error:Expected type 'Countable|array'. Found 'string'
After
count((array)$collection['colors'])
It works for me!
'vendor\laravel\framework\src\Illuminate\Database\Eloquent\Builder.php' to:
$originalWhereCount = is_array($query->wheres) ? count($query->wheres) : 0;
I;m using laravel 6.x
for this case you can use this way:
$id = \DB::table('xxxx')->where('id', $id)->count();
I Solve this in Laravel 5.6
// in controller
public function index()
{
$todos = Todo::all();
return view('todos.index')->with(['todos' => $todos]);
}
// in view page
#if(count($todos) > 0)
#foreach($todos as $todo)
<div class="well">
<h3>{{$todo->text}}</h3>
<span class="label label-danger">{{$todo->due}}</span>
</div>
#endforeach
#endif
I want some help with breadcrumb in codeigniter. It's my first experience with CodeIgniter and I don't have to clear enough how can I declare this function at Controller
public function breadcrumb(){
$this->load->library('breadcrumb');
function Tutorial() {
parent::Controller();
$this->load->library('breadcrumbcomponent');
$this->breadcrumbcomponent->add('Home', base_url());
$this->breadcrumbcomponent->add('Tutorials', base_url().'tutorials');
$this->breadcrumbcomponent->add('Spring Tutorial', base_url().'tutorials/spring-tutorials');
}
}
in View. I create the library as it is showing here:
http://www.technicalkeeda.com/codeigniter-tutorials/how-to-create-bread-crumb-using-php-codeigniter
Help me to understand better and to solve this case!
Best :)
Please edit your question and include proper code tags where needed. I will assume that you are using new version of CodeIgniter, but old tutorial (one that is related to CI version less than 2). Try to change this in code should be working that way.
In controller, change this code:
function Tutorial(){
parent::Controller();
$this->load->library('breadcrumbcomponent');
}
with this one:
public function __construct()
{
parent::__construct();
$this->load->library('breadcrumbcomponent');
}
Library class could stay the same, I'd say.
I am working on search filter on checkbox click, with Laravel and Ajax call. So I get results when I click on a checkbox. my query is as follows:
$editors = User::with(['editor.credentials','editor.specialties','editor.ratings']);
$temp=$editors->whereHas('editor', function($q) use ($a_data){
$q->whereHas('specialties',function($sq) use($a_data){
$sq->whereIn('specialty',$a_data);
});
})->paginate(2);
This gives me all the data I need. however how should I get the links for pagination?
$temp->setBaseUrl('editors');
$links = $temp->links()->render();
I am currently doing this and with $links which I am sending over as response to ajax call, I set the pagination with $links data. Now, I need to append the query to next page like page=2?query="something". I don't know how should I go about appending the remaining query result links to next page links. i.e. I don;t know what should come in the query="something" part. Can someone guide me. thanks
{{ $users->appends($_GET)->links() }}
It will append all query string parameters into pagination link
As of Laravel 7, you can call the withQueryString() method on your Paginator instance.
Quote from the documentation:
If you wish to append all current query string values to the pagination links you may use the withQueryString method:
{{ $users->withQueryString()->links() }}
See "Appending To Pagination Links": https://laravel.com/docs/7.x/pagination#displaying-pagination-results
Check the answer from #Arda, as it's global solution. Below you can find how to do it manually.
Use appends on Paginator:
$querystringArray = Input::only(['search','filter','order']); // sensible examples
// or:
$querystringArray = ['queryVar' => 'something', 'anotherVar' => 'something_else'];
$temp->appends($querystringArray);
Append all input except the actual page, form token and what you don't want to pass:
$paginatedCollection->appends($request->except(['page','_token']));
For the latest version of Laravel at the moment (5.2), you can just use the Request facade to retrieve the query string and pass that to your paginator's appends() method
$input = Request::input();
$myModelsPaginator = App\Models\MyModel::paginate();
$myModelsPaginator->appends($input);
Add this anywhere in your app (e.g routes.php, filters.php or anything that's autoloaded), no need to edit any pagination codes that is written already. This works flawlessly using view composers, and you don't need to know any query string parameters:
////////PAGINATION QUERY STRING APPEND
View::composer(Paginator::getViewName(), function($view) {
$queryString = array_except(Input::query(), Paginator::getPageName());
$view->paginator->appends($queryString);
});
//////////////////
Inspired from previous answers I ended up using the service container for both frontend + api support.
In your AppServiceProvider#boot() method:
$this->app->resolving(LengthAwarePaginator::class, function ($paginator) {
return $paginator->appends(array_except(Input::query(), $paginator->getPageName()));
});
you can used request helper in view as same
{{ $users->appends(request()->query())->links() }}
in your view where you display pagination...
{{ $results->appends(Request::except('page'))->links() }}
appends keeps the query string value except "page". not sure if it will work with POST request
{{ $users->appends(Request::only(['filter','search']))->links()}}
Updating #rasmus answer for Laravel 8.
In your AppServiceProvider boot method, add the following lines and your existing query string will be be used for all pagination links
$this->app->resolving(Paginator::class, function ($paginator) {
return $paginator->appends(Arr::except(Request::query(), $paginator->getPageName()));
});
$this->app->resolving(LengthAwarePaginator::class, function ($paginator) {
return $paginator->appends(Arr::except(Request::query(), $paginator->getPageName()));
});
And for completeness, use these classes:
use Illuminate\Pagination\LengthAwarePaginator;
use Illuminate\Pagination\Paginator;
use Illuminate\Support\Arr;
use Illuminate\Support\Facades\Request;
I'm developing a view that need to reuse a model, I'm following this documentation http://docs.joomla.org/Using_multiple_models_in_an_MVC_component. But that reference do the trick just (at least as far as I understand) when I use the parameter get task. if I use the view, joomla get me null data.
more clearly
controller.php - the task I named as the view I need
function viewdowhatIneed(){
$view = & $this->getView('viewdowhatIneed',html);
$view->setModel( $this->getModel( 'thenotdefaultmodelthatIneed' ), true );
$view->display();
}
model - thenotdefaultmodelthatIneed.php
class BLAModelthenotdefaultmodelthatIneed extends Jmodel{
function getReusableData0(){...}
function getReusableData1(){...}
}
view - view.html.php
class BLAViewviewdowhatIneed extends JView{
function display($tpl=null){
$dataneedit0 = $this->get('ReusableData0');
$dataneedit1 = $this->get('ReusableData1');
$this->assignRef('dataneedit0',$dataneedit0);
$this->assignRef('dataneedit1',$dataneedit1);
parent::display($tpl);
}
}
SO, what happen to me is:
example.com/index.php?option=com_BLA&view=viewdowhatIneed -> variables(datadataneedit0,dataneedit1) == NULL
example.com/index.php?option=com_BLA&task=viewdowhatIneed -> variables(datadataneedit0,dataneedit1) == Get me right data
then, my question is, is there a way to do the same thing, by using view parameter without task parameter (btw, I know this could be not a important problem, but I'm not an expert and on this reference http://docs.joomla.org/How_Joomla_pieces_work_together, it says:
The task part may or may not exist. Remember that if you omit it you are defaulting to task=display
so I really want to know that. In other words, can my view force to check the controller or vice versa.
Thanks in advance, excuse my english
I have problem with Joomla layouts in my component..There must be something bad with file/class name convention..
I was trying to ask on Joomla developer forum, but noone answered..
So I am getting this error.. 500 - View not found [name, type, prefix]: PostToBank,,postToBankView
The view is in views/postTobank/view.php and name is postToBankViewPostToBank
In my controllers/controller.php file I have class named PaymentController which contains this part of code.
$view = $this->getView('PostToBank','','postToBankView');
$this->getModel("Payment")->savePaymentData($data);
foreach ($data as $key => $value) {
$view->assignRef($key, $value);
}
$view->setLayout('postTobank');
$view->display();
my view.php file looks like this
class postToBankViewPostToBank extends JView{
function display($tpl=null){
//display set template
parent::display($tpl);
}
}
on attached image is full folder structure of my component..
Please whats wrong with this?Thanks
Joomla uses a naming conventions and you are not following them. Refer to http://docs.joomla.org/File_Structure_and_Naming_Conventions
Also, your views should be view.html.php and then you do not need to call setView.
FYI: this is where the error is coming from. Refer to this: http://docs.joomla.org/API16:JController/getView, even though it is 1.6 doc it is same in 1.5
Look though this tutorial and adopt regular conventions: http://docs.joomla.org/Developing_a_Model-View-Controller_Component_-_Part_1