Action App\Http\Controllers\AdminController#dbTanger not defined.
And this is My AdminController
public function dbTanger() {
$data = Db::table('ressos')
->where('secteur','Services')
->get();
return view('backend.layouts.admin.typeFilterTanger',compact('data','pagi'));
}
And this is the view
<div class="Filter">
<p>Filter Using Ville</p>
Tanger-Asilah
</div>
so if anyone can help me please
and thank you all
Since Laravel 9, the default Controller namespace has not been included in the RouteServiceProvider so you need to be explicit about where to locate controllers (or add the default namespace to your RouteServiceProvider).
What you can do is the following:
{{ action('\App\Http\Controllers\AdminController#dbTanger') }}
However, I would recommend using the route helper in conjunction with named routes as this is easier to manage should files change or move location in the future.
Route::get('/admin/dbTanger', [AdminController::class, 'dbTanger'])->name('admin.dbTanger');
Then use it as follows:
{{ route('admin.dbTanger') }}
The outcome is the same, just easier to manage and maintain long-term.
That error would mean you didn't define a route to this action so there is nothing in the route collection to be found for that action.
Define a route for this action and you would be able to use that action helper to create a URL to a route that uses this action.
I am using laravel 5.5. I have created a helper.php in app\Http. I am calling this helper in my blade file by using
{!! Helper::functionName() !!}
this is working fine. but i want to hold this helper result in a variable like
{!! $Result=Helper::functionName() !!}
But currently this is printing this result. How to solve this. please help.
So that i can make any if condition on this $Result.
In my helpers.php
namespace App\Http\Helpers;
class Helper
{
public static function functionName()
{
return "mydata";
}
}
There is no point to use helper like this. You should run the helper in controller and pass calculated data into view. In most cases you shouldn't set any variables in views or make any calculations - those should be passed from controller to view and view should only use them.
In this case, you can use "<?php ?>".
So result:
<?php $Result=Helper::functionName(); ?>
may be this is not possible because in laravel "{{}}" this means echo "" so by default it will print the value. return the value from helper function and use in your blade
I want to #include('item_1'), if not the item_1 view then #include('default').
Idk how to include default view.
My laravel version is 5.2.
You can try exists() method:
#include(View::exists('item_1') ? 'item_1' : 'default')
#if (View::exists('item_1'))
#include('item_1')
#else
#include('default')
Just use a if statement
I am using Laravel 4. I would like to access the current URL inside an #if condition in a view using the Laravel's Blade templating engine but I don't know how to do it.
I know that it can be done using something like <?php echo URL::current(); ?> but It's not possible inside an #if blade statement.
Any suggestions?
You can use: Request::url() to obtain the current URL, here is an example:
#if(Request::url() === 'your url here')
// code
#endif
Laravel offers a method to find out, whether the URL matches a pattern or not
if (Request::is('admin/*'))
{
// code
}
Check the related documentation to obtain different request information: http://laravel.com/docs/requests#request-information
You can also use Route::current()->getName() to check your route name.
Example: routes.php
Route::get('test', ['as'=>'testing', function() {
return View::make('test');
}]);
View:
#if(Route::current()->getName() == 'testing')
Hello This is testing
#endif
Maybe you should try this:
<li class="{{ Request::is('admin/dashboard') ? 'active' : '' }}">Dashboard</li>
To get current url in blade view you can use following,
Current Url
So as you can compare using following code,
#if (url()->current() == 'you url')
//stuff you want to perform
#endif
I'd do it this way:
#if (Request::path() == '/view')
// code
#endif
where '/view' is view name in routes.php.
This is helped to me for bootstrap active nav class in Laravel 5.2:
<li class="{{ Request::path() == '/' ? 'active' : '' }}">Home</li>
<li class="{{ Request::path() == 'about' ? 'active' : '' }}">About</li>
A little old but this works in L5:
<li class="{{ Request::is('mycategory/', '*') ? 'active' : ''}}">
This captures both /mycategory and /mycategory/slug
Laravel 5.4
Global functions
#if (request()->is('/'))
<p>Is homepage</p>
#endif
You can use this code to get current URL:
echo url()->current();
echo url()->full();
I get this from Laravel documents.
I personally wouldn't try grabbing it inside of the view. I'm not amazing at Laravel, but I would imagine you'd need to send your route to a controller, and then within the controller, pass the variable (via an array) into your view, using something like $url = Request::url();.
One way of doing it anyway.
EDIT: Actually look at the method above, probably a better way.
You will get the url by using the below code.
For Example your URL like https//www.example.com/testurl?test
echo url()->current();
Result : https//www.example.com/testurl
echo url()->full();
Result: https//www.example.com/testurl?test
For me this works best:
class="{{url()->current() == route('dashboard') ? 'bg-gray-900 text-white' : 'text-gray-300'}}"
A simple navbar with bootstrap can be done as:
<li class="{{ Request::is('user/profile')? 'active': '' }}">
Profile
</li>
The simplest way is to use: Request::url();
But here is a complex way:
URL::to('/').'/'.Route::getCurrentRoute()->getPath();
There are two ways to do that:
<li{!!(Request::is('your_url')) ? ' class="active"' : '' !!}>
or
<li #if(Request::is('your_url'))class="active"#endif>
You should try this:
<b class="{{ Request::is('admin/login') ? 'active' : '' }}">Login Account Details</b>
The simplest way is
<li class="{{ Request::is('contacts/*') ? 'active' : '' }}">Dashboard</li>
This colud capture the contacts/, contacts/create, contacts/edit...
For named routes, I use:
#if(url()->current() == route('routeName')) class="current" #endif
Set this code to applied automatically for each <li> + you need to using HTMLBuilder library in your Laravel project
<script type="text/javascript">
$(document).ready(function(){
$('.list-group a[href="/{{Request::path()}}"]').addClass('active');
});
</script>
instead of using the URL::path() to check your current path location, you may want to consider the Route::currentRouteName() so just in case you update your path, you don't need to explore all your pages to update the path name again.
In Blade file
#if (Request::is('companies'))
Companies name
#endif
class="nav-link {{ \Route::current()->getName() == 'panel' ? 'active' : ''}}"
Another way to write if and else in Laravel using path
<p class="#if(Request::is('path/anotherPath/*')) className #else anotherClassName #endif" >
</p>
Hope it helps
Try this:
#if(collect(explode('/',\Illuminate\Http\Request::capture()->url()))->last() === 'yourURL')
<li class="pull-right"><a class="intermitente"><i class="glyphicon glyphicon-alert"></i></a></li>
#endif
For Laravel 5.5 +
<a class="{{ Request::segment(1) == 'activities' ? 'is-active' : ''}}" href="#">
<span class="icon">
<i class="fas fa-list-ol"></i>
</span>
Activities
</a>
1. Check if URL = X
Simply - you need to check if URL is exactly like X and then you show something. In Controller:
if (request()->is('companies')) {
// show companies menu or something
}
In Blade file - almost identical:
#if (request()->is('companies'))
Companies menu
#endif
2. Check if URL contains X
A little more complicated example - method Request::is() allows a pattern parameter, like this:
if (request()->is('companies/*')) {
// will match URL /companies/999 or /companies/create
}
3. Check route by its name
As you probably know, every route can be assigned to a name, in routes/web.php file it looks something like this:
Route::get('/companies', function () {
return view('companies');
})->name('comp');
So how can you check if current route is 'comp'? Relatively easy:
if (\Route::current()->getName() == 'comp') {
// We are on a correct route!
}
4. Check by routes names
If you are using routes by names, you can check if request matches routes name.
if (request()->routeIs('companies.*')) {
// will match routes which name starts with companies.
}
Or
request()->route()->named('profile')
Will match route named profile. So these are four ways to check current URL or route.
source
#if(request()->path()=='/path/another_path/*')
#endif
Try This:
<li class="{{ Request::is('Dashboard') ? 'active' : '' }}">
<a href="{{ url('/Dashboard') }}">
<i class="fa fa-dashboard"></i> <span>Dashboard</span>
</a>
</li>
There are many way to achieve, one from them I use always
Request::url()
Try this way :
registration
I am using Eloquent together with Laravel 4's Pagination class.
Problem: When there are some GET parameters in the URL, eg: http://site.example/users?gender=female&body=hot, the pagination links produced only contain the page parameter and nothing else.
Blade Template
{{ $users->link() }}
There's a ->append() function for this, but when we don't know how many of the GET parameters are there, how can we use append() to include the other GET parameters in the paginated links without a whole chunk of if code messing up our blade template?
I think you should use this code in Laravel version 5+.
Also this will work not only with parameter page but also with any other parameter(s):
$users->appends(request()->input())->links();
Personally, I try to avoid using Facades as much as I can. Using global helper functions is less code and much elegant.
UPDATE:
Do not use Input Facade as it is deprecated in Laravel v6+
EDIT: Connor's comment with Mehdi's answer are required to make this work. Thanks to both for their clarifications.
->appends() can accept an array as a parameter, you could pass Input::except('page'), that should do the trick.
Example:
return view('manage/users', [
'users' => $users->appends(Input::except('page'))
]);
You could use
->appends(request()->query())
Example in the Controller:
$users = User::search()->order()->with('type:id,name')
->paginate(30)
->appends(request()->query());
return view('users.index', compact('users'));
Example in the View:
{{ $users->appends(request()->query())->links() }}
Be aware of the Input::all() , it will Include the previous ?page= values again and again in each page you open !
for example if you are in ?page=1 and you open the next page, it will open ?page=1&page=2 So the last value page takes will be the page you see ! not the page you want to see
Solution : use Input::except(array('page'))
Laravel 7.x and above has added new method to paginator:
->withQueryString()
So you can use it like:
{{ $users->withQueryString()->links() }}
For laravel below 7.x use:
{{ $users->appends(request()->query())->links() }}
Not append() but appends()
So, right answer is:
{!! $records->appends(Input::except('page'))->links() !!}
LARAVEL 5
The view must contain something like:
{!! $myItems->appends(Input::except('page'))->render() !!}
Use this construction, to keep all input params but page
{!! $myItems->appends(Request::capture()->except('page'))->render() !!}
Why?
1) you strip down everything that added to request like that
$request->request->add(['variable' => 123]);
2) you don't need $request as input parameter for the function
3) you are excluding "page"
PS) and it works for Laravel 5.1
In Your controller after pagination add withQueryString() like below
$post = Post::paginate(10)->withQueryString();
Include This In Your View
Page
$users->appends(Input::except('page'))
for who one in laravel 5 or greater
in blade:
{{ $table->appends(['id' => $something ])->links() }}
you can get the passed item with
$passed_item=$request->id;
test it with
dd($passed_item);
you must get $something value
In Laravel 7.x you can use it like this:
{{ $results->withQueryString()->links() }}
Pass the page number for pagination as well. Some thing like this
$currentPg = Input::get('page') ? Input::get('page') : '1';
$boards = Cache::remember('boards' . $currentPg, 60, function() {
return WhatEverModel::paginate(15);
});
Many solution here mention using Input...
Input has been removed in Laravel 6, 7, 8
Use Request instead.
Here's the blade statement that worked in my Laravel 8 project:
{{$data->appends(Request::except('page'))->links()}}
Where $data is the PHP object containing the paginated data.
Thanks to Alexandre Danault who pointed this out in this comment.