Till now i am using the form post method like this::
controller ::
public function loading_view()
{
.
.
.
.
$this->load->view('abc');
}
view abc:
">
when this form posted it will redirected to
public function method_of_controller()
{
.
//perform query
.
// here i am havin 2 way to call
//1.
$this ->load->view('abc');
//and 2.
load->loading_view();
}
//bu url remain same after the post method like ....index.php/loading_view
and but on reloading the page again the query runs fr 2nd method of loading view
so which way you will suggest me to use best
1- If you are not passing data to your view and your view is just contains html and javascript
than first method is beter.
2- if you are passing data to our abc view page than second is beter.
3- beter way is to make another view for this part for portability
because in some point you will reach you make changes so it is beter to each should be
separate.
yes you can post data to save view, without changing url redirection, in following way
controller
class PostTest extends CI_Controller{
public function loading_view()
{
$this->form_validation->rules('name','Name','required');
if(!$this->form_validation->run()){
$this->load->view('abc');
}
else{
$name = $this->input->post('name');
}
}
}
now here is most interesting part is view in form action you need to use uri_string() method like this
view
<? echo form_open(uri_string())
.form_input('name').br()
.form_error('name')
.form_submit('submit','Post data');
?>
Related
I searched more time to find how to get the controller method name by passing the URL but not found my expected answer. I want to make a method where I will pass a URL and it will give the corresponding controller action like as below but I can't figure out.
I found a helper which just return the current URL's action which is Route::currentRouteAction()
If a route in my application like as Route::get('/abc', 'YourController#method') which will generate the url http://example.com/abc
then how can I get the YourController#method by passing http://example.com/abc
function getAction($url){
//what will be logic?
// return like App\Controllers\MyController#method
}
I have to make a custom permission system where I need it for show and hide the menu by checking the URL of each menu.
Within your controller you can do the following:
<?php
use Illuminate\Routing\Router;
use Illuminate\Http\Request;
public function index(Request $request, Router $route)
{
$action = $router->getRoutes()->match($request)->getActionName();
// action should be what you're looking for.
}
You can try this if you want to:
Route::get('/the/url', 'YourController#method');
Every time anything calls the URL in the route, your method will be called.
You don't need to navigate to that url to call your method, it could be called by a form action, or a buttons action and just execute your method.
Edit:
url is your url as parameter (plain route)
import this:
use Illuminate\Routing\Route;
this is your function:
public function method(Route $route, $url)
{
$routes = \Route::getRoutes()->getRoutes();
foreach($routes as $r){
if($r->getUri() == $url){
$youraction= $r->getActionName();
dd($youraction);
}
else{
dd('does not exist');
}
}
}
Tested.
I'm trying to learn laravel 4. I created a form(using view) and returned it via a controller(testController) using index method. I had created this controller using artisan command.
i created another method (dologin) in the controller which would process the form. In the form url parameter i gave the address of dologin method.
This is the route:
Route::resource('test', 'testController');
This is the controller
<?php
class testController extends \BaseController {
public function index()
{
return View::make('test.index');
}
public function dologin(){
echo "working";
}
and this is the index view file
{{ Form::open(array('url'=>'test/loginform')) }}
{{ Form::text('username', null, array('placeholder'=>'Username')) }}<br/>
{{ Form::password('password', array('placeholder'=>'Password')) }}<br/>
{{ Form::submit('Login') }}
{{ Form::close() }}
After submitting form, it should echo "working" in the browser. But after submitting the form, page is blank. The url changes though from
/laravel/public/index.php/test/
to
/laravel/public/index.php/test/loginform
umefarooq's answer is correct, however hopefully this answer should give you a bit more insight into getting a head-start in your Laravel development as well as a consistent best-practice programming style.
Firstly, class names should really start with a capital letter. Try to keep methods / function names starting with a lower case letter, and class names starting with a capital.
Secondly, you don't need the \ in front of BaseController. You only need the backslash if you are name-spacing your controller. e.g. if your controller is in the folder Admin\TestController.php, and you put your TestController in the Admin namespace by typing <?php namespace Admin at the beginning of the file. This is when you should use \BaseController because you are telling your TestController to extend BaseController from the Global Namespace. Alternatively, before you declare your class, you can type use BaseController; and you don't need to put a \ in every time.
Specifically related to your question:
When you use resource routes in your routes file, you are telling Laravel that the controller can have any or all of the following methods: index, show, create, store, edit, update and destroy.
As such, Route::resource('test', 'TestController'); will point to TestController.php inside your controllers folder.
Your TestController should be structured as follows, most restful controllers will use the below as some kind of boilerplate:
<?php
class TestController extends BaseController
{
public function __construct()
{
}
// Typically used for listing all or filtered subset of items
public function index()
{
$tests = Test::all();
return View::make('test.index', compact('tests'));
}
// Typically shows a specific item detail
public function show($id)
{
$test = Test::find($id);
return View::make('test.show', compact('test'));
}
// Typically used to show the form which creates a new resource.
public function create()
{
return View::make('test.create');
}
// Handles the post request from the create form
public function store()
{
$test = new Test;
$test->attribute1 = Input::get('attribute1');
$test->attribute2 = Input::get('attribute2');
$test->attribute3 = Input::get('attribute3');
$test->attribute4 = Input::get('attribute4');
if ($test->save())
{
return Redirect::route('test.show', $test->id);
}
}
// Shows the edit form
public function edit($id)
{
$test = Test::find($id);
return View::make('test.edit', compact('test'));
}
// Handles storing the submitted PUT request from the edit form.
public function update($id)
{
$test = Test::find($id);
$test->attribute1 = Input::get('attribute1');
$test->attribute2 = Input::get('attribute2');
$test->attribute3 = Input::get('attribute3');
$test->attribute4 = Input::get('attribute4');
if ($test->save())
{
return Redirect::route('test.show', [$id]);
}
}
// Used to delete a resource.
public function destroy($id)
{
$test = Test::find($id);
$test->delete();
return Redirect::route('test.index');
}
}
Also, the beauty of using Resource Controllers is that you can take advantage of named routes.
in the terminal window, type in php artisan routes.
You should see 7 named routes.
test.index
test.destroy
test.show
test.edit
test.destroy
test.create
test.update
So within your form, instead of doing
{{ Form::open(array('url'=>'test/loginform')) }} you can point the url to a named route instead:
{{ Form::open(array('route' => array('test.store')) }}
That way if you ever change the url, or need to move around your site structure, this will be easy, because the forms post url will auto bind to the named route within the routes file. You wont need to update every single one of your views to ensure that the url's are pointing to the correct location.
Finally, as a starting point, I would recommend using JefreyWay/Laravel-4-Generators package. https://github.com/JeffreyWay/Laravel-4-Generators . Use them to create your resources, controllers, views etc. and see how the generators scaffold your models, views and, controllers for you.
Here is another resource to help you get started:
https://laracasts.com/lessons/understanding-rest
Route::resource('test', 'testController');
will work for RESTful method of controller, like index, edit, destroy, create and now you are using custom method of controller for this you need to create another route
Route::post("test/loginform",'testController#dologin');
hope this will work for you. read route documentation http://laravel.com/docs/routing
In addition to what umefarooq said, which is 100% accurate. You need to look into flash messages as well.
public function dologin(){
//do login verification stuff
If login validated
Return redirect::to(logged/page)->with('message', 'You're logged in');
If login failed
Return redirect::to('test')->with('message', 'You login credentials fail');
}
For further research:
http://laravel.com/docs/responses
I'm trying to generate ajax specific responses from my controllers by using the Request::ajax() method, which is working just fine. The only problem is that the way I have it set up right now isn't really a nice looking solution.
My controller:
class HomeController extends BaseController {
protected $layout = 'layouts/main';
public function __construct()
{
$this->beforeFilter('auth');
}
public function getIndex()
{
$view = View::make('content.home.index');
if(Request::ajax()) return $view; //For ajax calls we only want to return the content to be placed inside our container, without the layout
$this->layout->menu = 'content.menu';
$this->layout->content = $view;
}
}
So right now, for every method I define within my controllers I need to add the code snippet that checks for an AJAX request and returns a single view if the statement returns true.
This leads to my question that is probably more PHP related than it is to the framework;
Is there a way of executing my AJAX check on every method call, without actually placing it inside the method? Or is there some other solution to keep my code DRY?
Thanks in advance!
PS: This is my first post on stackoverflow, so feel free to correct me if I made any mistakes
Create a new barebone layout named 'layouts/ajax' (or any name you like).
<?php echo $content ?>
In your Base controller, override this setupLayout() function.
protected function setupLayout()
{
if ( ! is_null($this->layout))
{
$layout = Request::ajax() ? 'layouts/ajax' : $this->layout;
$this->layout = View::make($layout);
}
}
Change your getIndex() function to this.
public function getIndex()
{
$view = View::make('content.home.index');
$this->layout->menu = 'content.menu';
$this->layout->content = $view;
}
Now non-ajax requests will be rendered using layout set in the controller, where as ajax requests will receive whatever set to $this->layout->content.
Note : Controller will neglect the layout setup in setupLayout(), if the called method returns truthy value. So this method will not work for functions like below.
public function getIndex()
{
return View::make('content.home.index');
}
You could just change the layout property, in the constructor, if it's an ajax request:
public function __construct()
{
$this->beforeFilter('auth');
if(Request::ajax()) {
$this->layout = '';
}
}
If it doesn't work try setting it to NULL instead.
Why would you return a VIEW via ajax? Are you using it to create a SPA? If so there are better ways. I'm generally against returning HTML via AJAX.
The route I'd go in your position is probably opposite of how you're doing it. Render the view no matter what, if the request is ajax, pass the extra data back and have JS render the data on the page. That's essentially how most Javascript MVC frameworks function.
Sorry if I am totally missing the point here, just going on an assumption of your end goal with the info you provided.
I have a table of data from a database that I want to display on various pages of my website. Ideally, i would like to just have an include or something that will go off and get the data, and return the html table. The html and data will be identical everytime I need to use this.
I wanted to know the best way of doing this
Thanks
EDIT
If this helps, something similar to a Django "inclusion" custom tag...for any django developers reading
you need to pass the variable $data to the view method.
This is your code:
function load_my_view(){
$this->load->model('my_table');
$data['my_results'] = $this->my_table->my_data();
$this->load->view('my_view');
}
Please change it to this in order to load the $data into the view:
function load_my_view(){
$this->load->model('my_table');
$data['my_results'] = $this->my_table->my_data();
$this->load->view('my_view',$data);
}
You should use a function in a model to fetch the data you need. Your controller calls the model function and sends the returned information to a view. You don't need to use traditional php includes with Codeigniter. I recommend a review of the user guide. It's very good and will tell you all the basic stuff you need to know to develop with CI. But to get you started, you watn to use Models, Views, and Controllers. Your url will tell CI what controller and function inside that controller to run. If your url is
http://www.example.com/my_controller/load_my_view
Then CI will do what is inside the load_my_view function in the my_controller controller. function load_my_view in turn instantiates a model "my_table" and runs a database query, returns information that the controller sends to the view. A basic example follows:
Your model
class my_table extends CI_Model{
function my_data(){
$this->db->select('column_1,column_2,column_3');
$this->db->from('my_table');
$query = $this->db->get();
if($query->num_rows()>0){
$result = $query->result();
}
else{
$result = false;
}
return $result;
}
}
Your controller
class my_controller extends CI_Controller{
function load_my_view(){
$this->load->model('my_table');
$data['my_results'] = $this->my_table->my_data();
$this->load->view('my_view');
}
}
Your View
<ul id = "my_db_results">
<?php foreach($my_results as $result):?>
<li><?php echo $result->column_1." : ".$result->column_2." ( ".$result->column_3." )";?></li>
<?php endforeach;?>
</ul>
It looks like a good oportunity to use cache: http://codeigniter.com/user_guide/libraries/caching.html
Ok, so here is one thing that worked for me, but its definitely not perfect.
I created a view in which made a call to the model getting the data then put the data into a table. This way, I only have to include this view to put the table anywhere.
I understand this completely ruins the point of having a MVC framework, but hopefully it demonstrates what I want to do...and it works
I'm fairly new to Zend Framework and MVC in general so I'm looking for some advice. We have a base controller class in which we have some methods to obtain some user information, account configurations, etc.
So I'm using some of those methods to write out code in various controllers actions, but now I want to avoid duplicating this code and further more I would like to take this code outside of the controller and in a view helper as it is mainly to output some JavaScript. So the code in the controller would look like this:
$obj= new SomeModel ( $this->_getModelConfig () );
$states = $obj->fetchByUser ( $this->user->getId() );
//Fair amount of logic here using this result to prepare some javascript that should be sent to the view...
The $this->_getModelConfig and $this->user->getId() are things that I could do in the controller, now my question is what is the best way to pass that information to the view helper once i move this code out of the controller ?
Should I just call these methods in the controller and store the results into the view and have the helper pick it up from there ?
Another option I was thinking of was to add some parameters to the helper and if the parameters are passed then I store them in properties of the helper and return, and when called without passing the parameters it performs the work. So it would look like this:
From controller:
$this->view->myHelper($this->user->getId(), $this->_getModelConfig());
From view:
<?= $this->myHelper(); %>
Helper:
class Zend_View_Helper_MyHelper extends Zend_View_Helper_Abstract
{
public $userId = '';
public $config = null;
public function myHelper ($userId = null, $config = null)
{
if ($userId) {
$this->userId = $userId;
$this->config = $config;
} else {
//do the work
$obj = new SomeModel($this->config);
$states = $obj->fetchByUser($this->userId);
//do the work here
}
return $this;
}
}
Any advice is welcomed!
Firstly the ASP style ending tag here at "$this->myHelper(); %>" is bad practice, with that said it is more advisable to keep the logic in the model and the controller just being used to call the model, get the results and spit that to the view for viewing.
what I would do is, if i simply want to pass a bunch of values to the view, i stuff them in an associative array and send them over.
anyway you should not be doing your ...
"//Fair amount of logic here using this result to prepare some javascript that should be sent to the view..."
part in the controller, I would advice you to make a new model that does that logic stuff for you, and you just call your model in the controller pass it what ever arguments that are needed and then spit the result of that to the view.
The best way is to get the data from your model throught your controller, and then pass to the view. But if you really need a custom helper to echo the view parts, we only will know if you say exactly what you're trying to do.
If you already have this logic in a helper, try to just pass the parameters in your view myhelper($this->params); ?>
You may want take a look at this approach too:
// In your view to put javascript in the header
// You can loop trought your data and then use it to generate the javascript.
<?php $this->headScript()->captureStart(); ?>
$().ready(function(){
$('#slideshow').cycle({
fx: 'fade',
speed: 1000,
timeout: 6500,
pager: '#nav'
});
});
<?php $this->headScript()->captureEnd() ?>