I am new in Laravel using version 5.8
I do not want to set route manually for every controller.
What i want is that if i give any url for example -
www.example.com/product/product/add/1/2/3
www.example.com/customer/customer/edit/1/2
www.example.com/category/category/view/1
for the above example url i want that url should be treated like
www.example.com/directoryname/controllername/methodname/can have any number of parameter
I have lots of controller in my project so i want this pattern should be automatically identified by route and i do not need to specify manually again and again Directory Name, Controller , method and number of arguments(parameter) in route.
try this:
Route::get('/product/edit/{id}',[
'uses' => 'productController#edit',
'as'=>'product.edit'
]);
Route::get('/products',[
'uses' => 'productController#index',
'as'=>'products'
]);
in the controller:
public function edit($id)
{
$product=Product::find($id);
return view('edit')->with('product',$product);
}
public function index()
{
$products=Product::all();
return view('index')->with('products',$products);
}
in the index view
#foreach($products as $product)
Edit
#endforeach
in the edit view
<p>$product->name</p>
<p>$product->price</p>
Related
I'm developing an application with Laravel 5.8. In my application, I have a controller that handles backend articles, and it works. I want to display my user-side information in such a way that a user can click on a link and see the detail of an article. For that, I have created a new controller a with a new namespace for the function show my redirection of navigation in different page does not focus that it is en route or URL with Laravel 5.8. Below is the function.
namespace App\Http\Controllers\Cybernaut;
use App\History;
use App\Http\Controllers\Controller;
class HistoryController extends Controller
{
public function show($id)
{
$history = History::find($id);
return view('show_history', compact('history'));
}
}
At the level of the home page I wanted to have my links like these:
<li><a data-hover="history" href="{{route('history.show',$history→id)}}"><span>history</span></a></li>
Error
ErrorException (E_ERROR) Property [id] does not exist on this
collection instance. (View:
C:\laragon\www\venome\resources\views\layouts\partial\header.blade.php)
And here is the route used for the show function.
Route::group(['namespace'=>'cybernaut'], function (){
Route::get('/history/{slug}','HistoryController#show')->name('history.show');
});
Try after modifying the thing I have these at the route level now.
Route::get('/', 'FrontController#index')->name('index');
Auth::routes();
Route::get('/home', 'HomeController#index')->name('home');
Route::get('/admin/dashboard', 'DashboardController#index')->name('admin.dashboard');
Route::group([], function () {
Route::get('/history', 'HistoryController#index')->name('history.index');
Route::get('/history', 'HistoryController#create')->name('history.create');
Route::get('/history/edit', 'HistoryController#update')->name('history.update');
Route::get('/history', 'HistoryController#destroy')->name('history.destroy');
});
Route::group(['namespace' => 'cybernaut'], function () {
Route::get('/history/{history}', [
'as' => 'show',
'uses' => 'HistoryController#show'
]);
});
At the level of the homepage I wanted to put my link like those here now;
#foreach($history as $history)
<li><a data-hover="history" href="{{url('/history/'.$history->id)}}"><span>history</span></a></li>
#endforeach
I have this error now:
Trying to get property 'id' of non-object (View:
C:\laragon\www\venome\resources\views\layouts\partial\header.blade.php)
I want an internaut to be able to navigate between the pages.
You have a conflict of variables on your homepage.
#foreach($history as $history)
should be
#foreach($histories as $history)
where $histories is filled in in your FrontController.
$histories = History::all();
When actually getting your single history object, I agree with Sapnesh's answer that you best doublecheck whether or not the object actually exists.
The error occurs because find() returns NULL when a model is not found.
find($id) takes an id and returns a single model. If no matching model
exist, it returns null.
findOrFail($id) takes an id and returns a single model. If no matching
model exists, it throws an error.
In your show() method,
Change:
$history = History::find($id);
To:
$history = History::findOrFail($id);
Let's say I have a function in my controller which retrieves users looking something like this:
public function index($category) {
// retrieve users depending on category or all
}
Now is there a way to make named routes to include the function parameter like so:
Route::get('passengers', 'Controller#index(1)')->name('passengers');
Route::get('attendees', 'Controller#index(2)')->name('attendees');
This way they can all use the same function
No you can not pass a parameter the action name, and there is a problem in you routing logic :
Route::get('/{categoryName}', 'Controller#index')->name('index');
And in the controller you will for example get the category by name like this :
public function index($categoryName) {
$category = Category::where('name', $categoryName)->first();
// use $ category as you please ;)
}
In the blade :
route('index', ['categoryName' => $category->name])
If the named route defines parameters, you may pass the parameters as the second argument to the route function. The given parameters will automatically be inserted into the URL in their correct positions
https://laravel.com/docs/5.5/routing#named-routes
So, use route() helper like this:
route('passengers', ['category' => 1])
Then you need to add {category} to the route. Also, it's really better to use show() instead of index() here. So, your route will look like this:
Route::get('passengers/{category}', ['as' => 'passengers', 'uses' => 'Controller#show']);
Yes, you can define the param in the url like so:
Route::get('passengers/{yourParam}', 'Controller#index')->name('passengers');
View in docs
Route::get( '{category}', [ 'as' => 'users', 'uses' => 'Controller#index' ]);
Remember to add this route at the end of your routes file in order to not to collide with any other route.
Now in your controller
use Illuminate\Http\Request;
public function index(Request $request)
{
$category = $request->query('category');
// $category will be passengers, attendees, etc
}
Your routes will be
/passengers can be accessed as route('users', ['category' => 'passengers'])
/attendees can be accessed as `route('users', ['category' => 'attendees'])
i have translated url which i need to redirect to the specific controller function, but i need also to pass an exact parameter.
For example i want to show all football news, but in the url i do not have the ID of the sport football (id=1) so i need to pass the parameter id=1 to the index() function.
Route::get('/football-news/', ['as' => 'news.index', 'uses' => 'NewsController#index']);
it is not an option to pass 'football' as a parameter, because it is just an example. The real route is translated and the code looks like that:
Route::get(LaravelLocalization::transRoute('routes.football.news'), ['as' => 'news.index', 'uses' => 'NewsController#index']);
suppose you have a NewsController to fetch all news be like
class NewsController extends Controller
{
public function index()
{
$news = News::all(); //you have to create News model
return view('news.index', compact('news')); //use to pass data in view
}
public function show($id)
{
$news_detail=News::find($id); //to fetch detail of news from database
return view('news.show', compact('news_detail'));
}
}
create index.php and show.php in views/news folder. in index.php
#foreach($news as $news_item)
<div>
{{ $news_item->title }}
</div>
#endforeach
here using "/news/{{$news_item->id}}" you can pass id of specific news into route file.
in show.php
<h1>news</h1>
<h1>
{{ $news_detail->title }}
</h1>
<ul class="list-group">
#foreach($news_detail->detail as $details)
<li class="list-group-item">{{$details}}</li>
#endforeach
</ul>
in route file
Route::get('/news/{news}', 'NewsController#show');
now you have to create show($id) function in NewsController.php which parameter is id.
You can append the index URL with ?id=1 parameter (eg. domain.com?id=1) and get it in your index controller action by using Request::get('id');
For example:
Url in template file:
<a href="domain.com?id=1" />
In your NewsController:
public function index(Request $request){
$id = $request->get('id');
}
You should be able to have access to the parameter even though you didn't specify wildcards in the route file.
Edit:
you will have to call a different #action for a different route. You can pass in an id wildcard.
For example, in Route file:
Route::get('tennis-news/{id}', 'NewsController#tennisIndex');
Route::get('football-news/{id}', 'NewsController#footballIndex');
Then in the NewsControlleryou must have public methods tennisIndex($id) and footballIindex($id), these methods will have access to the wildcard you set in the route.
For example, in NewsController
public function tennisIndex($id){
$tennnis_news = News::where('sport'='tennis)->where('id', $id)->get();
return view('tennis_news', compact('tennnis_news'));
}
I like to use resource controllers in Laravel, as it makes me think when it comes to data modelling. Up to now I’ve got by, but I’m now working on a website that has a public front-end and a protected back-end (administration area).
I’ve created a route group which adds an “admin” prefix, like so:
Route::group(array('before' => 'auth', 'prefix' => 'admin'), function()
{
Route::resource('article', 'ArticleController');
Route::resource('event', 'EventController');
Route::resource('user', 'UserController');
});
And I can access the methods using the default URL structure, i.e. http://example.com/admin/article/1/edit.
However, I wish to use a different URL structure on the front-end, that doesn’t fit into what resource controllers expect.
For example, to access an article, I’d like to use a URL like: http://example.com/news/2014/06/17/some-article-slug. If this article has an ID of 1, it should (under the hood) go to /article/1/show.
How can I achieve this in Laravel? In there some sort of pre-processing I can do on routes to match dates and slugs to an article ID, and then pass that as a parameter to my resource controller’s show() method?
Re-visiting this, I solved it by using route–model binding and a pattern:
$year = '[12][0-9]{3}';
$month = '0[1-9]|1[012]';
$day = '0[1-9]|[12][0-9]|3[01]';
$slug = '[a-z0-9\-]+';
// Pattern to match date and slug, including spaces
$date_slug = sprintf('(%04d)\/(%02d)\/(%02d)\/(%s)', $year, $month, $day, $slug);
Route::pattern('article_slug', $date_slug);
// Perform the route–model binding
Route::bind('article_slug', function ($slug) {
return Article::findByDateAndSlug($date_slug);
});
// The actual route
Route::get('news/{article_slug}', 'ArticleController#show');
This then injects an Article model instance into my controller action as desired.
One simple solution would be to create one more route for your requirement and do the processing there to link it to the main route. So, for example:
//routes.php
Route::get('/arical/{date}/indentifier/{slug}', array (
'uses' => 'ArticleController#findArticle'
));
//ArticleContoller
public function findArticle($date,$slug){
$article = Article::where('slug','=','something')->first(); //maybe some more processing;
$article_id = $article->id;
/*
Redirect to a new route or load the view accordingly
*/
}
Hope this is useful.
It seems like if Laravel 4 supports (:all) in routing, you would be able to do it with ease, but unfortunately (:all) is not supported in Laravel 4.
However, Laravel 4 allows detecting routes by regular expression, so we can use ->where('slug', '.*').
routes.php: (bottom of the file)
Route::get('{slug}', 'ArticleController#showBySlug')->where('slug', '.*');
Since Laravel will try to match the top most route in routes.php first, we can safely put our wildcard route at the bottom of routes.php so that it is checked only after all other criteria are already evaluated.
ArticleController.php:
class ArticleController extends BaseController
{
public function showBySlug($slug)
{
// Slug lookup. I'm assuming the slug is an attribute in the model.
$article_id = Article::where('slug', '=', $slug)->pluck('id');
// This is the last route, throw standard 404 if slug is not found.
if (!$article_id) {
App::abort(404);
}
// Call the controller's show() method with the found id.
return $this->show($article_id);
}
public function show($id)
{
// Your resource controller's show() code goes here.
}
}
The code above assumes that you store the whole URI as the slug. Of course, you can always tailor showBySlug() to support a more advanced slug checking.
Extra:
You could also do:
Route::get('{category}/{year}/{slug}', 'ArticleController#showBySlug')->where('slug', '.*');
And your showBySlug() would just have additional parameters:
public function showBySlug($category, $year, $slug)
{
// code
}
Obviously you can extend to month and day, or other adaptations.
I have a multilingue website created using Laravel 4, and I have lot of pages such as : "policy, "terms", "how it works" in database, so to access thoses pages I use this route:
// Group by locale
Route::group(
array( 'prefix' => $locale ), function () {
Route::get('{slug}', array('uses' => 'PageController#show','as' => 'pages.show');
// Website routes
});
And then I search for the given slug and the current locale.
My is problem is that I can't add for example a page link in the footer because the slug is dynamic. so is there any solution to resolve that.
It's make a sense ?
Thanks
You are already catching the slug in
Route::get('{slug}', array('uses' => 'PageController#show','as' => 'pages.show');
part. all you need is to inject this slug into controller like this:
class PageController extends BaseController {
public function show($slug)
{
return 'showing slug ' . $slug;
}
}
and whatever value the route receive for {slug} part in route laravel will automatically inject that value into the controller.