If I dd($items); in the controller, the result like this :
In the view blade laravel I check if collection empty like this :
#if($items)
...
#endif
But it does not work
How can I solve this problem?
You could use $items->isEmpty(); or $items->isNotEmpty();
Like so:
#if(!$items->isNotEmpty())
...
#endif
You can further read over here:
https://laravel.com/docs/5.5/collections#method-isempty
i normally double check so that if i don't get the value the page keeps running :
#if(isset($items))
#if(!empty($items))
.....
#endif
#endif
hope it helps
#if ( $items->count() )
....
#endif
Related
Theres any Solution to hide an Code in Laravel for one Page?
With this i can display only one requested page ,Example:
#if (Request::path() == 'message/send')
#endif
Theres an opposite to hide an code for an path?
I have an URL also with after send an Username:
message/send/username
I have try: #if (Request::path() == 'message/send/') and #if (Request::path() == 'message/send/{{$username}}') but dont work.
Thanks
Use request()->is() for this:
#if (request()->is('message/send*'))
you can use segment method in laravel
#php $member = Request::segment(1);#endphp
#if($member =='member')
<style type="text/css"> .test{color: #1a393a !important;</style>
#endif
#endif
Here my first segment is member
The controller is the proper place to handle request specifics. However, if you still wish to do this in your blade view:
#if (Request::path() == 'message/send/' . $username)
You do not use template syntax {{ }} inside expressions. When the blade template is compiled, #if (expr) is converted to <?php if (expr): ?>, so standard PHP applies within the expression.
Route path comparison doesn't work that way. For example:
message/send/jwz104 will never be equal to message/send/{{$username}}
You could compare the route name:
#if(\Request::route()->getName() == 'message.send')
#endif
This also isn't a nice solution. You should handle this in your controller or by view composers.
with core php-
#if (strpos($_SERVER['REQUEST_URI'], "message/send") !== false)
#endif
I am trying to set active classes on active menus. In the past I was using Request::is() function for this, but in the new version of laravel it says "Class 'Request' not found."
As you are using blade you can use request helper method. Try like this
request()->is('your_url');
do not put / before the route.
it should be like
#if(Request::is('index'))
#include('include.showcase')
#endif
please check in your code
you add this line as namespace or not
use Illuminate\Http\Request;
and you can also pass one varriable with view from controller side and check on blade template if you get that data then you can use active class for that menu. like this
in controller what i had used in my project
$data['page-tab'] = 'adduser';
return view('user.add_user',compact('data'));
and in blade file i had checked this like this in sidebar section
<li class="{{ isset($data['page-tab']) && $data['page-tab'] == 'adduser' ? 'active' : '' }}"><i class="fa fa-file-text-o"></i>Manage Member</li>
hope this works for you.
Use it without the '/'
like this:
#Request::is('home')
not
#request::is('/home')
I am trying to retrieve data from database and check if the data is empty or not. What problem I am facing is that html is showing even if the data is empty. I want to ignore the html tag like example ul li. Here how i tried is like
#if(!empty($jobseekers->skill_1))
<li> My Skill is : {{ \App\skill::where('id',$jobseekers->skill_1)->pluck('name')->first() }}</li><br/>
#endif
I want to ignore "My Skill is " if the data is empty. I don't want to show anything.
When using ->get() you can't simply use any of the below:
if (empty($jobseekers->skill_1)) { }
if (!$jobseekers->skill_1) { }
if ($jobseekers->skill_1) { }
But, When you are getting data with first() method, You can simply use all above methods.
Because if you dd($jobseekers->skill_1); you'll notice an instance of Illuminate\Support\Collection is always returned, even when there are no results.
I think you are using !empty() on data with ->get() method which will always return true even data is empty. You need to use other way.
To determine if there are any results you can do any of the following:
if (!$jobseekers->skill_1->isEmpty()) { }
if ($jobseekers->skill_1->count()) { }
if (count($jobseekers->skill_1)) { }
If you get $jobseekers with get() method you can not use empty($jobseekers )
instead of empty you can use other conditions :
#if($jobseekers->skill_1 != '')
in this condition you check skill_1 as empty string
also
#if($jobseekers->skill_1)
and etc
replace your code with below code and check it:
#if($jobseekers->skill_1 != '')
<li> My Skill is : {{ \App\skill::where('id',$jobseekers->skill_1)-pluck('name')->first() }}</li><br/>
#endif
you should use isset()
#if(isset($jobseekers->skill_1))
<li> My Skill is : {{ \App\skill::where('id',$jobseekers->skill_1)->pluck('name')->first() }}</li><br/>
#endif
you can use count method
#if(count($jobseekers->skill_1)>0)
<li> My Skill is : {{ \App\skill::where('id',$jobseekers->skill_1)-pluck('name')->first() }}</li><br/>
#endif
#if(count($data)>0)
//write your code which you want to show if data is available
#endif
I have a custom view and in some functions, I used paginate and other functions I don't use paginate. now how can I check if I used paginate or not ?
#if($products->links())
{{$products->links()}}
#endif // not work
of course that I know I can use a variable as true false to will check it, But is there any native function to check it ?
This works perfectly. Check if $products is an instance of Illuminate\Pagination\LengthAwarePaginator then display the pagination links.
#if($products instanceof \Illuminate\Pagination\LengthAwarePaginator )
{{$products->links()}}
#endif
#if($threads->hasPages())
{{ $threads->links() }}
#endif
Simple one!
Try like this
#if($products instanceof \Illuminate\Pagination\AbstractPaginator)
{{$products->links()}}
#endif
You need to check wheater the $products is instance of Illuminate\Pagination\AbstractPaginator. It can be an array or Laravel's Collection as well.
As of laravel 7 you can now do this:
#if( $vehicles->hasPages() )
{{ $vehicles->links() }}
#endif
The beautiful way:
#if ($products->hasMorePages())
{{ $products->links() }}
#endif
Click here to see the official documentation
Don't use a check on the base class of the variable. This could lead to problems with changing base classes in future Laravel versions. Simply check whether the method links exists:
#if(method_exists($products, 'links'))
{{ $products->links() }}
#endif
Corrected code (add "isset"):
#if(isset($products->links()))
{{$products->links()}}
#endif
Shorter version:
{{$products->links() ?? ''}}
It will work for paginate, simplePaginate and when there is no pagination. The solution with "$products->hasMorePages()" will not display pagination links on the last page.
Another way:
#if (class_basename($products) !== 'Collection')
{{ $products->links() }}
#endif
You can use PHP function: get_class($products) - to get full class name.
Laravel should have some function to check ->paginate() is in use.
just write paginate(0) instead of get()
Blade Template: simply use {{$products->links}}. no #if #endif needed.
laravel paginate have 2 type :
simplePaginate() will return \Illuminate\Pagination\Paginator
paginate() will return Illuminate\Pagination\LengthAwarePaginator
Based on the above conditions, you can try this solution :
#if(
$products instanceof \Illuminate\Pagination\Paginator ||
$products instanceof Illuminate\Pagination\LengthAwarePaginator
)
{{ $products->links() }}
#endif
Juse use below format
#if($products instanceof \Illuminate\Pagination\LengthAwarePaginator )
{{$products->links()}}
#endif
#if($products->currentPage() > 1)
{{$products->links()}}
#endif
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