Laravel equivalent or alternative for CodeIgniter - php

I am trying to figure out how to do the equivalent of the following in Laravel that I would do in CodeIgniter all the time to build views:
$section = $this->load->view('pages/about', $data, TRUE);
This would allow me to echo $section in another view file and then when that view was called the normal way, it would render it. I am not sure how to do something like this in Laravel.
UPDATE
I figured it out. What I was needing was Laravel's HtmlString class to take a string and convert it to html markup to the view file.

You would need to use the View Facade, so make sure to include it with an "Use" statement in your Controller, but basically is this:
$html = View::make('pages/about', $data)->render();
The render() method will just render the view in HTML, instead of returning it as a Response object like the view() helper function does.

There are several ways to do so, try this:
return view('admin.profile', $data);
Read through this doc:
https://laravel.com/docs/5.5/views

Related

Rendering data in Laravel 5.1 using shortcodes like wp

I have the below content in my DB-
<p>This is dummy content for testing</p>
{{LandingPageController::getTest()}}
I want to render that into my view. But when I'm rendering this in Laravel view, this {{LandingPageController::getTest()}} is getting displayed as it is stored in DB.
I want to call the LandingPageController getTest method in my view.
Please suggest me a quick fix for this.
Landing Page Controller
public function getTest(){
return "Hello World!!!";
}
just make the function static
public static function getTest(){
return "Hello World!!!";
}
that's the only way you can call it like this {{LandingPageController::getTest()}} but I do advice not to do that in your blade file this not a good code design. you should do $test = LandingPageController::getTest() in the controller that you return the blade view and pass it like this return view('blade_file_name',compact('test')) and in your blade file just do {{$test}}
PS - if you doing it your controller use the class like this use Path\To\Controller\LandingPageController
Use namespace for that controller in your blade file. example
namespace App\Http\Controllers\LandingPageController;
You can evaluate a string as a php code using the eval() function
eval — Evaluate a string as PHP code
But it is highly discouraged.
The eval() language construct is very dangerous because it allows execution of arbitrary PHP code. Its use thus is discouraged. If you have carefully verified that there is no other option than to use this construct, pay special attention not to pass any user provided data into it without properly validating it beforehand.
You can use a generic string, {test} for example, when saving the content in the storage.
<p>This is dummy content for testing</p>
{test}
Then whenever you need to display the actual content, you can simply replace the generic string with the real value. You'll have this line in your blade file:
{{ str_replace('{str}', "Hello World", $content) }}
Take a look at Helper. You can call helper function in view to render your text or html
Got the solution, achieve the functionality with "laravel-shortcodes".
Found a very good tutorial on laravel-shortcodes like wordpress

Get data from model to every view using OOP and DRY

I have a piece of data I want passed to every view. I am using CodeIgniter 3 and have PHP 7 available to me. The current way I do it is using something like this in every function.
$data['foobar'] = $this->general_model->foobar();
// More code
$this->load->view('homepage', $data);
I'd prefer not to have to call $data['foobar'] = $this->general_model->foobar(); on every single function.
I've tried many approaches to fix this without resorting to anything that makes the code too goofy. I've tried constructors, autoload, and hooks. The problem in each case boils down to the fact that $data is local to each function. The best I've gotten is usually something like this.
$data['foobar'] = $this->foobar;
// More code
$this->load->view('homepage', $data);
This is slightly nicer, but it still results in me placing this line in every function.
I'd like my functions to in someway inherit $data with the index foobar already set. I'd prefer to avoid a solution that requires every function receiving $data as a parameter. How can I accomplish this?
Option 1:
Not sure if you have tried this but you could set $data as a property of your class
protected $data = [];
Then in your constructor set it.
$this->data['foobar'] = $this->general_model->foobar();
This would mean your $data becomes accessible to all your methods in your controller and you would need to refer to them as $this->data['data_name'] and use it in a view like
$this->load->view('homepage', $this->data);
Option 2:
A second way is to create a method like render() which is common to all your methods that load views and replaces your existing view calls.
So you would have something like...
public function one_of_my_methods(){
$data['content'] = 'This is content 1';
$this->render('test_view',$data); // Call the new view handler
}
// All methods using views now call this to load the final view
public function render($view,$data){
$data['foobar'] = 'I am common'; // DRY
$this->load->view($view, $data);
}

PHP Phalcon: no views?

I'm evaluating frameworks for use with an API, and I'm taking a serious look at PHP Phalcon. It's looking promising - "lean" (load what you need), but with a lot of options.
I'm wondering... is it possible to not use views (templates, rather) with it? Do I have to set up a view, or can I just output .json?
Thanks!
There is a way in Phalcon to disable view in the action and avoid unnecessary processing:
public function indexAction() {
$this->view->disable();
$this->response->setContentType('application/json');
echo json_encode($your_data);
}
Depending on what you want to do you can disable the view as others have suggested and echo json encoded data, or you could use the built in response object as below:
$this->view->setRenderLevel(View::LEVEL_NO_RENDER);
$this->response->setContentType('application/json', 'UTF-8');
$this->response->setJsonContent($data); //where data is an array containing what you want
return $this->response;
There is also a tutorial in the docs that goes over building a simple REST api
http://docs.phalconphp.com/en/latest/reference/tutorial-rest.html
If you won't be using any views at all you can disable views at the very start.
$app = new \Phalcon\Mvc\Application();
$app->useImplicitView(false);
Even if you do this, I think you still have to set the view DI for the framework to work.
Also, if you want to output json there's a method for that:
$this->response->setJsonContent($dataArray);
$this->response->send();
Yeah, you can do it, I'm using PHP Phalcon.
To ignore view, in your controller your action should be like
public function indexAction() {
$var = array or other data
die(json_encode($var));
}
die(); in controller will not render any parent layout! :)

How to simulate a request from view.

I'm new in cakephp and I'm just wondering, how to test models and controllers without using views?
I have to simulate saving data using models and controllers without using froms from views. I was thinking about to make an array with the needed values, but maybe there is a better way to do that?
you can mock your model functions using code like:
$model = $this->getMockForModel('MyModel', array('save'));
$model->expects($this->once())
->method('save')
->will($this->returnValue(true));
You can output variables at any time from controllers (or models) without getting to the views. Yes, it's not how you should do things with an MVC framework, but for testing, it's pretty easy to whack this below your database call in the model/controller:
<? echo '<pre>'; print_r($my_array); exit; ?>
The other thing you can do is at the top of your action function in the controller put:
$this->layout = '';
$this->render(false);
... which will bypass the layout and skip the view rendering, so you can output whatever you like within that function without using the view.
At the beginning of your action, you may use:
$this->autoRender = false;
This will allow you to access your action directly by going to it's path (e.g. CONTROLLER/ACTION). Before passing your data array to save() or saveAll(), I recommend double-checking it with Debugger::dump(), and follow that with die(). This will make the array containing the save data print on your screen so you can verify it looks proper and follows Cake's conventions. The die() will prevent it from actually saving the data.
If everything looks correct, remove the dump() and die() and test it out again.
The first response, from Ayo Akinyemi, should also work well if you are Unit Testing your application.

How to define function for multiple urls in Codeigniter

I want my Codeigniter to call a function export(...) when
domain.com/foo/bar/show/122.export
also the same function if
domain.com/boo/for/show_all/172.export
domain.com/goo/par/detail/122.export
so I thought I could define that function export(...) only once
and pass the HTML code of the respective page to the export function,
or at least the source of the call.
How can I define this for my whole CI-page?
you can use CI's URI route
something like this should work (say your export function is in exports controller)
$route['([a-z]+)/([a-z]+)/([a-z]+)/(\d+)\.(export)'] = "exports/export/$1/$2/$3/$4";
So your export method will have source information.

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