I have generated the default Laravel auth module.
Everywhere in the blades of the module, I see Double Underscore __ function assuming that translation is almost there.
for example
<li>
<a class="nav-link" href="{{ route('login') }}">
{{ __('Login') }}
</a>
</li>
My question: Where is the translation file? Where should I put it, if I create one?
I mean, if I go to the Laravel documentation site there are examples like this
echo __('messages.welcome');
with explanations
For example, let's retrieve the welcome translation string from the resources/lang/messages.php language file:
BUT in my example above there is no file name specified. It is only text:
__('Login')
Question is: What is used for the language file if no file specified? Is there any default? Where does it sit? Where was it set?
All the language translation files in Laravel should be stored in PROJECT_DIRECTORY/resources/lang. When you make an Auth with artisan, it automatically creates it. But if you can't find it, then create manually.
(1)
There's a way to using translation strings as keys by the docs. In this method you can create a JSON file in PROJECT_DIRECTORY/resources/lang with the name of your local, for example for Spanish name it es.json or German de.json, it depends on your local name.
Now create a JSON object and put the translations with the string name you used in your blade:
{
"Login": "Welcome to Login Page!",
"Logout": "You are logged out!",
}
Then use the PHP double underscores method to call your translations in blades:
{{ __('Login') }}
(2)
Create a file named auth.php in PROJECT_DIRECTORY/resources/lang directory. then put a simple php array like this on it:
<?php
return [
/*
Translations go here...
*/
];`
Then add your translate strings to it:
<?php
return [
'Login' => 'Welcome to Login Page!',
'Logout' => 'You are logged out!',
];`
Now in the blade template simply do this:
<li>
<a class="nav-link" href="{{ route('login') }}">
{{ __('auth.Login') }}
</a>
</li>
Laravel Docs
Have an instruction about the json file. Yes it is not php, but json file. Example would be:
resources/lang/es.json
content
{
"I love programming.": "Me encanta programar."
}
Usage
echo __('I love programming.');
It looks like there are no translation file for the default __('Login'), __('Register'), ... provided by laravel.
By default if no translation is found for __('foobar'), laravel just uses the string in the parentheses. So here, assuming there is no translation file, __('Login') is expanded to 'Login'.
I have a simple problem, basically I am getting name of the website from database and create a link according to it's name. it looks like:
#foreach ($websites as $website)
<a class="websites" href=" {{ asset ($website->name )}}"> {{ asset ($website->name )}}
</a>
#endforeach
Which gives for example: http://localhost/name
Howver links needs to be like this:
http://localhost/website/name how can I add /website into my URL using blade template in laravel?
Try this:
{{ url('website/' . $website->name) }}
This have some improvement on #Laran answer regarding best practices.
You would better use url parameters instead of concatenating the $name parameter
{{ url('website', [$name]) }}
And using named routes will be better to decouple the routing from the views.
// routes/web.php
Route::get('website')->name('website');
and write inside your {{ route('website', [$name]) }}
I'm using laravel and trying to delete something. Is it possible to specify the DELETE method on laravel's route()??
e.g
route('dashboard-delete-user', ['id' => $use->id, 'method'=> 'delete'])
or something like that??
EDIT:
What I meant was could I specify that in a link or a button in my blade template. Similar to this:
href="{{ route('dashboard-delete-user') }}
Yes, you can do this:
Route::delete($uri, $callback);
https://laravel.com/docs/master/routing#basic-routing
Update
If for some reason you want to use route only (without a controller), you can use closure, something like:
Route::get('delete-user/{id}', function ($id) {
App\User::destroy($id);
return 'User '.$id.' deleted';
});
No or at least I haven't figure out how to.
The only way for this to work out of the box would be to build a form to handle it. At the very minimum, you would need...
<form action="{{ route('dashboard-delete-user') }}" method="POST">
{{ method_field('DELETE') }}
{{ csrf_field() }}
<button type="submit" value="submit">Submit</button>
</form>
Or you can just create the get route which you are trying to link to and have it handle the logic. It doesn't need to be a route which only respondes to delete requests to delete a resource.
Yes you can, using a URL helper. https://laravel.com/docs/5.2/helpers#urls
There are several options to choose from.
Using a view with user input. I then want to pass to a route. This what I found so far:
href="{{URL::to('customers/single'$params')}}"
I want to pass the user input as the above $params to my route. This is sample of my route:
Route::get('customer/{id}', function($id) {
$customer = Customer::find($id);
return View::make('customers/single')
->with('customer', $customer);
As soon as I can pass the parameter I can do what I want with the route, which I know how.
Basically you can pass parameter to routes by doing:
Route::get('user/{name}', function($name)
{
//
})
->where('name', '[A-Za-z]+');
In your anchor tag, instead of doing href={{URL...}}, do something like:
{{ URL::to('user/$param') }}
For more information on routing, visit link
This is what I have and works:
<a <button type="button" class="buttonSmall" id="customerView" href="{{URL::to('customer',array('id'=>'abf'))}}" >View</button></a>
But I need the array value 'abf' to be the value of a textbox.
This worked for me in my view anchor tag
href="{{ URL::to('user/'.$param) }}"
instead of what was specified above
href="{{ URL::to('user/$param') }}"
You can user it in View as I used:
<a class="stocks_list" href="/profile/{{ Auth::user()->username }}">Profile</a>
Hope it helps you.
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