How to pass an array to controler from view ?? - laravel-5

Hi guys I have an array that is created in my View side and I want to pass it to a contoller so I can insert it in to my database.. I have seen examples that use json encode and decoding but how do i apply this to my laravel 5 project ?

That depends on your controller code. However, one way for that is to pass them as query parameters.
Link

Related

defining routes with parameters in laravel

in Laravel i want to do a page with a search box and a form (the route could be /products)
I want to retrieve information using search box typing id from a db and populate the form.
I request datas with the route for example /products/{id}
But in the controller i use the same function products($Request request) with if the id exists do something if no do other things, or there are two different functions?
Thx
Please go through the Laravel Resource Controllers.
To show list of products i.e. /products, create a index() method. To show a specific product i.e. /product/{id}, create show() method.
Probably is better use the same function, if id exist, return the page with the values filling the form, if not returns the same page but with a message showing that product doesn't exist

How to get columns of a model instance in Laravel?

Can I extract a specific column for a model instance in Laravel instead of using query methods statically? more clearly, instead of writing the code below:
MyModel::select('columnName')->where('id', $model->id)->first();
can we write?
$model->select('columnName');
Well, when you grab it from a model class, it is a Eloquent Object, and it has all attributes from the database.
But if you are talking about serializing that into an array or JSON everytime, then you need to modify your $visible array ;
if you wanna transform it for a specific case, then you need to do :
MyModel::select('columnName')->where('id', $model->id)->first()->pluck('columnName');

Cakephp 3.6. How to call a helpers query from a controller?

I have on my homepage a small search form with a list of categories.
I use a helper to query the database to get all categories. This is fine and works great.
Now I want to update the categories select with an updated list of categories based on a performed search.
That means, I would need to call query inside the helper with a given $variable to run a new query and to return new data.
But this is my concern now.
I can't call a helper from within the PagesController, right?
So what is the right way to
- have a helper call to use on the fly in all views selecting DB Data?
- address an ajax call to a controller and to use the query from the helper?
Any advice will be highly appreciated!
Thanks!
Short answer:
You code needs to be moved in a Table class and not in a Helper class, if you need it both in the Controller and the View.
Long Answer:
The code you have created in your helper needs to be moved in the relevant Table class and that will fetch the data.
Then in both your view and the controller you will use the new method you created in your Table class.
For example in your CategoriesTable class you can add a method called:
public function fetchCategoryData(){
//TODO: Custom code that will fetch the category data like...
//return $this->find();
}
public function updateCategoyData($args){
//TODO: Custom code to update your category data
}
Now, in either the View or the Controller you can use this
$Categories = TableRegistry::get('Categories');
$Categories->fetchCategoryData();
//or
$Categories->updateCategoryData();

Where to process data in Laravel

Back when I was using CodeIgniter I had functions in my models like
public function GetArticlesFormatted($inactive = FALSE)
and then in my controller I could have had
$articles->GetArticlesFormatted(true);
And so on.
How should I achieve the same with Laravel 5.4? The database of the app I'm building is already built and full and is a mess so most of the "automated" super-restrictive things that Laravel has don't work out of the box.
For example there is a Country Code that I'm retrieving and I need it as is, but in some instances I need it converted in a Country Name which I had in another table.
Right now I just have in my controller wherever I am retrieving data from the model:
$countryResult = Country::where('country_code', $item['country_code'])->first();
$countryArray = $countryResult->toArray();
$item['country'] = $countryArray['country_name'];
Can I somehow achieve that in a more elegant way?
I tried accessors but for some reason couldn't get anything to work for my purposes.
A select query can be used to limit selection to a particular column.
$countryName = Country::select('country_name')
->where('country_code', $item['country_code'])
->first();

Yii CGridView pagination

I'm using CArrayDataProvider (which is basically a customized query i've created) that returns all the results (over 1000) from the database.
I'm using the results in the view but when i'm using the pagination it's going back to the controller for another query.
my question is: is there any way to move on the the next set of results (already part of the result array) without going to the controller and model again.
*My controller has a fairly advanced function which requires variables and parameters which i dont have in the view when trying to use standard AJAX request for the next page.
thanks,
Danny
my question is: is there any way to move on the the next set of
results (already part of the result array) without going to the
controller and model again
Then my answer would be NO if you was using CGridview's pagination. In your situation, you have to make the pagination by yourself instead. You have already selected all of records, and would like to manipulate them on your client side, you really don't need the pagination of CGridview at all.
Pushing all of records into a page on first load is not good idea, but maybe your requirement has asked, I just say that.

Resources