How to use custom pagination in laravel 5.2 using Illuminate\Pagination\LengthAwarePaginator , i think this gives more flexibility than other class according to thier doc. But couldn't figure it out how to do that . So i need help on this . Kindly guide me .
Code in controller :
public function index(Request $request )
{
$data=array();
//**** fetch data starts
$countries_db = DB::table('location_country as lc');
$countries_db=$countries_db->select(DB::raw('lc.id,lc.country_name,lc.country_3_code,lc.country_2_code,lc.lat,lc.lng,lc.published'));
$pagi_country=$countries_db;
//**** fetch data ends
//***** pagination code starts
$pagi_country = $pagi_country->paginate(3);
$pagi_country->setPath('country/');
$data['pagi_country']=$pagi_country;
$data['useinPagiAr']=$useinPagiAr;
//***** pagination code ends
return view('admin.country.countrylist', $data);
}
Code in View
we will get data from this variable $pagi_country passing it to foreach loop , and printing data. To print the pagination link following code is needed
echo $pagi_country->appends($useinPagiAr)->render();
Related
I want get id value from eloquent laravel and passing to variable to insert process. this is my eloquent to get data from siswa table
$siswa = Siswa::where('nis', $row['nis'])->first();
but when accessing id value like $siswa->id I'm getting error
enter image description here
Use this code
if($siswa){
$id=$siswa->id;
}
$siswa variable can be null. Look at its contents with the dd($siswa) command.
To escape the error you can use it like this:
if($siswa)
{
// $siswa not null.
}
You have two ways to fix this.
1/ You can use ->firstOrFail() to display a 404 page in case of your data is not found :
$siswa = Siswa::where('nis', $row['nis'])->firstOrFail();
// if $siswa is NULL, then it will display a 404 page, else it will continue
2/ You can add a simple condition to continue with your data :
$siswa = Siswa::where('nis', $row['nis'])->first();
if ($siswa) {
// then continue
}
First you can $id pass on your public function like
public function show($id)
{
$siswa = Siswa::where('id', $id)->first();
}
You can pass your id on your route
Im using Laravel v7 and i have a question about pagination.
So far im using 2 routes, 1rst to return a view with all rows from database, and 2nd receives an input and returns that view with the rows filtered by that input value.
But im using pagination, and on the 2nd route, when i try to go to 2nd page it gives me an error:
Symfony\Component\HttpKernel\Exception\MethodNotAllowedHttpException
The GET method is not supported for this route. Supported methods: POST.
I've tried to change my form method to GET but i need that the token doesn't appear on the page URL and beside that, when i go to 2nd page, it returns all the rows again.
Thats my code so far:
Routes:
Route::get('concessions', 'ConcessionController#index')->name('concessions.index');
Route::post('concessions/search', 'ConcessionController#search')->name('concessions.search');
Controller
class ConcessionController extends Controller
{
public function index()
{
$concessions = DB::table('concessions')->paginate(12);
return view('admin.concessions.index', compact('concessions'));
}
public function search(Request $request)
{
$name = $request->name;
$concessions = Concession::where('name', 'like', '%' . $name . '%')->paginate(12);
return view('admin.concessions.index', compact('concessions', 'name'));
}
}
Any way to do that?
Laravel pagination only works with get parameters.
You should use GET method for your search page. POST requests aren't meant for the purpose of displaying data. Why? There are many reasons, but to be short I will give you three examples :
1. When you access the first page, you get the data by GET request,
not POST request. So if you want to use POST request, you
need to access the page as POST request by sending data with
POST method.
2. With GET parameters, let's say you are on 5th page - you can
copy the link and paste it to friend and he will be able to view
the same content as you. With POST this is impossible.
3. You can not use back button with POST requests, if you manage to
get pagination to work.
POST requests are useful when you need to submit data to the server, in order to create new record, for example.
So I suggest you to change your route type to GET.
From my perspective, if you change your route code similar to below code, it will work properly with both methods of GET and POST.
Route::any('concessions/search', 'ConcessionController#search')->name('concessions.search');
I am new to laravel and I am trying to get a pagination function into my result pages, so I have the following function to generate results from query and I would like to have a pagination on the results page, but I don't seem to get it work correctly
public function showResults()
{
$selectedquery = Input::get('Annonces');
$what = Input::get('what');
$where = Input::get('where');
$results = DB::table('annonces')->where($selectedquery,'LIKE', '%'.$what.'%')
->where('Lieu','LIKE', '%'.$where.'%')
->get();
return View::make('results',array('results' => $results));
}
Any Help?
Well, for one, you're missing the call to ->paginate(n). Right now, your closure is ->get(), which returns all results for your annonces table. This is good, but doesn't work for pagination. Change the function like so:
$results = DB::table('annonces')->where($selectedquery,'LIKE', '%'.$what.'%')
->where('Lieu','LIKE', '%'.$where.'%')
->paginate(10);
This will return all results grouped into 10 results per page. Feel free to change that as you see fit.
Lastly, somewhere on your view where you display the results, you will need to use this code to display a page-viewer:
<?php echo $results->links(); ?>
<!-- OR -->
{{ $results->links(); }}
Also, be sure to check out the docs on Laravel's pagination. You'll find it's pretty comprehensive!
Laravel Pagination
Hope that helps!
I want to grab some data from a database and display on a layout page, I've basically started building a small CMS to get into Laravel and all has gone fine so far but now i'm at a wall, and can't find a solution.
I have a layout blade file like so: http://paste.laravel.com/1fB1 nothing majot but you will see i have used $page->meta_title etc in there and in my controller i have:
public function home()
{
$pages = Pages::all();
return View::make('frontend/home')->with('pages',$pages);
}
Which I have a pages model doing nothing else really like so:
class Pages extends Eloquent {
protected $table = 'pages';
}
So why is it trying to get property of non-object and I don't really want to use a foreach because this is going to be the frontend of my 'test' website so a foreach wouldn't suite.
You'll need to access these items as a multi-dimensional array if you don't want to loop through them.
$pages[0]['field_name_here']
or
$pages[1]['field_name_here']
Its a bit of a tough one to answer without knowing how you want your CMS to work.
For example, you could have a route as {pagename} in your routes.php file, then have a page controller where you would get the requested route from the variable passed in. This would then load the page you wanted using the variable
public function page( $pagename ) {
$page = Page::where('page_title', '=', $pagename)->first();
View::make('frontend/page', array( 'page' => $page ));
}
Using a route like that, and the controller, in your view you could use {{ $page->content }} to get the content of the requested page from the database and display it.
Hope this helps.
Edit: Example Route:
Route::get('{pagename}', 'PageController#page');
I am using codeigniter. From a model i have fetched some data in the controller using the following codes:
$data['english'] = $this->question_model->getRandomQuestions('english', 2);
So I know to access the data from view i would use the following code..
foreach($english as $eng){
echo $eng['columnName'];
}
But I want to travase the data in the controller doing some of my calculation. But how can I do that. I used the save above code in controller but it is then showing error.
The error message is as below :
Undefined variable: english
How can used those obtained data in controller using foreach loop. Any help would be appreciated...
thanx
Since the response is added in $data['english'] in your controller, then you would do:
foreach( $data['english'] as $i => $eng ) {
$data['english'][$i]['columnName'] = 'change with whatever';
}