Hello and thanks for your help. I've done some research and tried a few options but can't seem get this to work properly. I'm passing a URL with a query string into a function that loads the page via URL passed. However, I'm trying to find a way to paginate the results as well. Is there a way I can pass the query string url to Laravel's pagination links? Thanks.
My URL with query string
<a id="searchData"class="btn btn-primary ml-1 mr-5 text-light" title="Search" type="submit"
onclick="ajaxLoad('{{url('dma/data')}}?startDate='+$('#startDate').val()+'&endDate='+$('#endDate').val()
+ '&dmaNameFilter=' + encodeURI(dma_name) + '&memberNameFilter=' + encodeURI(member_name))">Search Results
</a>
I tried this for the links():
{{ $data->appends(request()->query())->links() }}
I have this in my Controller:
$data = Report::where('CallDate', '>=', $start_date)->where('CallDate', '<=', $end_date)->paginate(13)->appends(request()->query());
You can also add this:
$this->app->resolving(\Illuminate\Pagination\LengthAwarePaginator::class, function ($paginator) {
return $paginator->appends(Arr::except(request()->query(), $paginator->getPageName()));
});
To your AppServiceProvider
Try This
$data->appends(request()->input())->links()
You can pass any data to pagination by calling
{{ $paginator->links('view.name', ['foo' => 'bar']) }}
on your situation I think you want to pass query string to paginator; you may try
{{ $paginator->links('view.name', request()->getQueryString() ) }}
If you need to append querystrings for your ajax controller you'd better check https://github.com/spatie/laravel-query-builder
Since Laravel 8, You can simply use
$paginator->withQueryString();
Related
First of all, I would like to say that my English is not so good and is my first time posting here, so excuse me if I did something wrong! Well, I am starting practicing with Laravel and I am trying to create a URL for users can like posts. my URL and controller is right now this Route::post('/post/{post}/like', [LikeController::class, 'postLike'])->name('post.like'); where post is the posts id that i am trying to pass through my form action attribute. Here is my form:
#props(['id'])
<div {{ $attributes->merge(['class' => 'card-footer d-flex justify-content-around']) }}>
<form action= "{{ route('post.like' , $id) }}" method="post" >
#csrf
<button>submit</button>
</form>
</div>
If you wonder if the id is not actually passed into the component, I checked {{ dd($id) }} and it printed it.
My controller is this which it doesn't actually do anything right now. I am just trying to pass the id:
class LikeController extends Controller
{
public function postLike(Post $post) {
dd($post);
}
}
After all this i am getting the error:
Missing required parameter for [Route: post.like] [URI: post/{id}/like] [Missing parameter: id]. (View: blog\resources\views\components\post\interaction.blade.php)
I am having two days to find the problem and I am still trying this moment I am sending this... I can't found where is the mistake! If you could help me would be much appreciated! Ty in advance
You need to pass the ID as an associative array in your route().
{{ route('post.like', ['id' => $id]) }}
If the named route defines parameters, you may pass the parameters as the second argument to the route function. The given parameters will automatically be inserted into the generated URL in their correct positions and you should name 'post' instead of 'id':
Route::post('/post/{post}/like', [LikeController::class, 'postLike'])->name('post.like');
And you can call this route anywhere
route('post.like', [$postId])
If that still doesn't work for you. Then it might be issue of Route Cache Run:
php artisan route:clear
Use same name parameter
When you use "id" keyword as a parameter in web.php then also same name pass in function argument in the controller
Route::post('/post/{id}/like', [LikeController::class, 'postLike'])->name('post.like');
Controller
class LikeController extends Controller
{
public function postLike(Post $id) {
dd($id);
}
}
Sorry for my bad English.
I have no idea this could be the problem but after many changes i cast the id of the post into an integer and it actually worked <form action= "{{ route('post.like' , (int)$id) }}" method="post" > . I don't know why i should cast the id into an integer because its already an integer by default and i checked that with {{ dd(getType($id)) }} and printed "integer"! I am really confused but it works! But i am not happy because i have no idea why i should cast an integer into an integer to make it work! Doesn't make any sense! If anyone has an answer to this would be great!
#foreach($users as $user)
$users[] = $user->name;
#endforeach
I want to get the output like this.
['X', 'Y', 'Z', 'C'];
Note: i need this because i have to use this in JavaScript function.
First of all, you can't "turn a foreach loop into an array", you would use that foreach loop to get an array as outcome. In this case I wouldn't go with a foreach approach, there are some methods that can help you cleaning out your code a bit.
You have two options depending on the type of the $users variable, pluck() or array_column():
// $users is a collection
json_encode($users->pluck('name'));
// $users is an array
json_encode(array_column($users, 'name'));
The json_encode() is highly recommended (as the comments pointed out) if you are going to use that output in your javascript. Now you can just send it in a respond and use it as a normal JSON.
In case you print the resulting variable using Blade, remember you need to use {!! !!}, otherwise, if you use {{ }} you would get unwanted scaped characters, since it uses htmlspecialchars() function under the hood.
Hope this helps you.
I recommend You should perform this action in your controller only, and then pass it to your view as
$userNames=implode(",",array_column($users,'name'));
return view('your_view',['userNames'=>$userNames]);
Then split it in your js as
var names="{!! $userNames!!}";
var names= names.split(",");
Collections will automatically be converted to json when you try to echo them.
The reason you're seeing " is because using {{ }} will automatically escape strings for your protection. You can prevent this by using {!! !!} instead.
Try something like this:
<script>
var data = '{!! $users->pluck('name') !!}';
console.log( JSON.parse(data));
</script>
Hope this helps!
I have a blog and a quotationfamous sayings repository on one site.
The quotations are tagged and the entries are tagged too.
I use this rtconner/laravel-tagging package.
Now, what I want to do is to display ALL Quotation models which share the same tags as article.
The Eloquent syntax is simple, as the original docs provide an example:
Article::withAnyTag(['Gardening','Cooking'])->get();
possible solution
Optional routing parameters. The asker-picked answer in this question gives a solution:
//in routes.php
Route::get('/{book?}/{chapter?}/{topic?}/{article?}', 'controller#func');
//in your controller
public function func($book = null, $chapter = null, $topic = null, $article = null) {
...
}
my problem
In my app the shared tags might count more than 3 or 5. I will soon get an example with even 10 tags. Possibly more
My question
Does it mean that I have to construct an URL with 10 optional routing parameters? Do I really need sth like this:
Route::get('quotations/tags/{tag1?}/{tag2?}/{tag3?}/{tag4?}/{tag5?}/{tag6?}/{tag7?}', 'controller#func');
my question rephrased
I could create a form with only a button visible, and in a hidden select field I could put all the tags. The route would be a POST type then and it would work. But this solution is not URL-based.
I think you could process the slashes, as data:
Route::get('quotations/tags/{tagsData?}', 'controller#func')
->where('tagsData', '(.*)');
Controller:
public function controller($tagsData = null)
{
if($tagsData)
{
//process
}
}
Ok, this is my solution. As I have a tagged model, I dont't need to iterate through tags in url to get the whole list of tags.
The enough is this:
// Routes file:
Route::get('quotations/all-tags-in/{itemtype}/{modelid}', 'QuotationsController#all_tagged_in_model');
Then in my controller:
public function all_tagged_in_topic($itemtype, $id) {
if($itemtype == 'topic') {
$tags = Topic::find($id)->tags->pluck('name')->all();
$topic = Topic::find($id);
}
if($itemtype == 'quotation') {
$tags = Quotation::find($id)->tags->pluck('name')->all();
$quotation = Quotation::find($id);
}
// dd($tags);
$object = Quotation::withAnyTag($tags)->paginate(100);;
And it is done.
Now, the last issue is to show tags in the URL.
TO do that, the URL should have an extra OPTIONAL parameter tags:
// Routes file:
Route::get('quotations/all-tags-in/{itemtype}/{modelid}/{tags?}', 'QuotationsController#all_tagged_in_model');
And in the {url?} part you can just write anything which won't break the pattern accepted by route definition.
In your view you might generate an URL like this:
// A button to show quotes with the same set of tags as the article
// generated by iteration through `$o->tags`
<?php
$manual_slug = 'tag1-tag2-tag3-tag4`;
?>
<a href="{{ URL::to('quotations/all-tags-in/article/'.$o->id.'/'.$manual_slug) }}" class="btn btn-danger btn-sm" target="_blank">
<i class="fa fa-tags icon"></i> Tagi:
</a>
Using a view with user input. I then want to pass to a route. This what I found so far:
href="{{URL::to('customers/single'$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 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.