active menu, but using id instead just use url laravel 9 - laravel

i want to show a link that is active using Request::is. if i just use url like Request::is('home') ? 'active' : null; that is no problem, but if i add id attribute, it didn't work like <a class="{{ Request::is('*#' . Str::slug($item->sub_title, '_')) ? 'text-sc underline font-semibold' : null }}" href="#{{ Str::slug($item->sub_title, '_') }}">{{ $item->sub_title }}</a>
so how to make Request::is() work with "#"

Related

Passing two parametres in url in octobercms

I must update a existing Octobercms project and it has url like this:
<a class="page-link" href="{{ 'catalog/category/' | page({'category': category.slug'}) }}" >Все товары</a>
And url in category page is this:
url = "/catalog/:category?"
How can I pass two parametres in url I tried this:
<a class="page-link" href="{{ 'catalog/category/' | page({'category': category.slug, 'second_parameter':'1'}) }}" >Все товары</a> and changed url in category page to url = "/catalog/:category?/:second_parameter?
but it did not help? Who can help me?
The important thing is the page's Filename name, as shown in the screenshot
Use it and create a link as shown below, it will work.
{{ 'test-pg'|page({'param1': 'one', 'param2': 'two'}) }}
Output
http://localhost:8000/test-pg/one/two
if any doubt please comment.

Is there a way to remove the %20 in Laravel URL?

Here's what I have so far.
in blade template
<a href='{{ url("/businessprofile/$business->id/$business->name") }}'>
and in web.php
Route::get('/businessprofile/{id}/{name}', 'BusinessController#show')
it shows
localhost:8000/businessprofile/User%20Info
is there a way to remove the %20 and just show localhost:8000/businessprofile/UserInfo instead?
The Str::slug method generates a URL friendly "slug" from the given string :
{{ url("/businessprofile/$business->id"."/" . Str::slug($business->name)) }}'>}}
Or,
{{ url("/businessprofile/$business->id"."/" . str_slug($business->name)) }}
If above method not work, then change your route & view as :
route :
Route::get('/businessprofile/{id}/{name}', 'BusinessController#show')->name('businessprofile.show');
view :
{{ route('businessprofile.show', ['id' => $business->id, 'name' => str_slug($business->name) ]) }}
See official documentation here
No, you should not do that. That's url encoding and ‰20 is code for space ( ).
change your url as below.
<a href="{{ url('businessprofile/'.$business->id.'/'.$business->name) }}">
Thank you for sharing your answers. I've tried mixing everyone's answer but the best I got is this
<a href='{{ url("/businessprofile/$business->id"."/".Str::slug($business->name)) }}'>
and it shows
http://localhost:8000/businessprofile/1/user-info

Laravel blade creating url

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]) }}

How to check for route namespace in Laravel

How to check for route namespace in Laravel similar to how the following code does?
{{ (Request::is('user*') ? 'active ' : '')
This way you can get a request segment:
{{ (Request::segment(1) == 'user') ? 'active' : '' }}
So /user/username, /user/show or /user etc, would all match.
http://laravel.com/docs/4.2/requests#request-information

How to Get the Current URL Inside #if Statement (Blade) in Laravel 4?

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

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