Laravel Blade: Use commands from string - laravel

I have variable in Controller:
$abc = '#include('partials.formerror', array(.....))';
And I send that variable to view:
\View::share([
"abc" => $abc,
]);
And I want in view:
......
......
#include('partials.formerror', array(....)
......
......
#include('partials.formerror', array(....) is dynamic content from Controller. But it's COMMAND of Blade, not plain text. How can I do that?

You need a string blade compiler like this one:
https://packagist.org/packages/wpb/string-blade-compiler
Laravel doens't allow blade rendering from strings out of the box.
Fyi: you could also use Blade::compileString which imho is not an elegant solution

your answer is here: https://laravel.com/docs/5.2/views#view-composers
You can create a view composer and attache dynamic content to your view.

Related

Laravel: how to create a rendered view from a string instead of a blade file?

I've some html with {{ soimething }} placeholders in a table.
I would like to get the rendered view from this custom html.
I would like to avoid to manually do string replace.
Is it possible?
Note : I seen suggested questions but I ended finding a more concise way to reach my goal. So I post an answer to this question. Please keep this open.
You can use Blade Facade.
use Illuminate\Support\Facades\Blade;
use Illuminate\Support\Facades\Blade;
public function __invoke()
{
$name='Peter Pan';
return Blade::render("
<h1> Hello {$name} </h1>
",['name'=>$name]);
}
Found
We can use \Illuminate\View\Compilers\BladeCompiler::render($string, $data)
Where
$string is the text to parse, for example
Hi {{$username}}
$data is the same associate array we could normally pass down to view() helper, for example [ 'username' => $this->email ]
I was missing this from the official doc: https://laravel.com/docs/9.x/blade#rendering-inline-blade-templates
So we can also use
use Illuminate\Support\Facades\Blade;
Blade::render($string, $data)

How can I pass a variable from slot to its layout in Laravel?

I am using Laravel 8 and I can't find a way to solve this problem.
I have a app-layout like this:
<!-- ./app-layout.blade.php -->
<html>
<head>...</head>
<body>
<livewire:create-suggestion />
{{ $slot }}
</body>
</html>
And this view uses the app-layout:
<!-- ./index.blade.php -->
<x-app-layout>
...
</x-app-layout>
In controller, I pass a variable to index view:
public function index() {
return view('index', ['variable', '$variable']);
}
How can I pass this variable to the app-layout? Because I also want to
use this in the create-suggestion livewire component.
Try this. edit your index.blade.php
#extends('app-layout', ['variable' => $variable])
...
You can pass any variable attribute to the layout like so:
<x-layout :myvariable=$myvariable>
or static attribute values like so
<x-layout myvariable="This is a string">
and retrieve it in the layout.blade.php like so:
{{ $attributes['myvariable'] }}
First off I beleive the App Layout will have access to whatever is passed by the view function. But if I am wrong or you wish to pass any custom variable whatsoever using the Layout syntaxt as opposed to template inheritance the way would be: (I came cross your question having a problem passing a static string for my HTML page title and no solution worked until I found this)
<x-app-layout>
<x-slot:variable>{{$variable}}</x-slot>
</x-app-layout>
However your issue is you are using Livewire and livewire is not simply just a view! Laravel Intentionally makes Livewire independent from the rest of the page. You would have to either pass it in PHP as param during mount which will only be accessible if set during the mount function() in \App\HTTP\LiveWire\createSuggestion.php
#livewire("create-suggestion",[ $variable])
Or ideally you have your javascript use the $emit feature to pass the element to the livewire listener all handled via Livewire. It seems to me you are using Livewire for something that you should simply only use Blade/Views:
#include("create-suggestion");

how to add common variable for laravel and pass this data to all view all views

I want to create a global variable and pass this into all view, so I can get this variable into all blade template
basically, my need is to pass my general setting controller value into my common blade view like header.blade.php
Thanks in advance
For that you will need to add some code in the App->Providers->AppServiceProvider.php like this:
public function boot()
{
view()->composer('*',function($view){
$settings = Settings::firstOrNew(['id' => '1']);
$view->with('settings', $settings );
});
}
According to Laravel documentation you can use view composer:
https://laravel.com/docs/5.8/views#view-composers
for using this feature in app > AppServiceProvider and in boot method you can use this approach for passing parameter to specific view If you have data that you want to be bound to a view each time that view is rendered. this approach meet your need perfectly. for example you want to pass a parameter named userName to header.
View()->composer('header', function ($view){
$userName= "username"
$view->with(['userName'=>$userName]);
});

How to call helper function in laravel 5.5

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

Laravel Pagination links not including other GET parameters

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.

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