in my controller in my show function in laravel i want the get the id that shows in browser show when i browse it it shows like this
http://localhost:8000/admin/invoices/1
i want to get that "1" and use it in show controller like below
public function show(Invoice $invoice)
{
$clients = Invoice::with('user','products')->get();
$invoice_id = 1;
$invoices = Invoice::with('products')->where('id', '=', $invoice_id)->firstOrFail();
return view('admin.invoices.show', compact('invoice','invoices'),compact('clients'));
}
and put it instead of $invoice_id so when every my client visit this page only sees the related invoice products . thanks you for help
If you're actually getting an instance of Invoice passed to your show method then it likely means you have Route-Model Binding set up for your project. Laravel is looking at the defined route and working out that the ID part (1) should map to an instance of Invoice and is doing the work to grab the record from the database for you.
The Invoice object passed through should refer to an item in your database with the ID of 1, so to get the ID that was mapped in the route you can simply just do:
public function show(Invoice $invoice)
{
echo $invoice->id; // This should be 1
Laravel supports route model binding out of the box these days, but in earlier versions you had to set it up in app/Providers/RouteServiceProvider.php. If you don't want it, try replacing your show method signature with this:
public function show($id)
{
echo $id; // Should be 1
By removing the type-hint you're simply expecting the value that was given in the route parameter and Laravel won't try to resolve it out of the database for you.
Simple way you may try this.
//Define query string in route
Route::get('admin/invoice/{id}','ControllerName#show')
//Get `id` in show function
public function show(Invoice $invoice,$id)
{
$invoice_id = $id;
}
Try using $invoiceId
public function show(Invoice $invoice, $invoiceId)
{
$clients = Invoice::with('user','products')->get();
$invoices = Invoice::with('products')->findOrFail($invoiceId);
return view('admin.invoices.show', compact('invoice','invoices'),compact('clients'));
}
do this if you want to get the url segment in controller.
$invoice_id = request()->segment(3);
if you want this in view
{{ Request::segment(3) }}
Goodluck!
Usually happens when giving a route name different from the controller name
Example:
Route::resource('xyzs', 'AbcController');
Expected:
Route::resource('abcs', 'AbcController');
Related
guys i trying to filter my data from mysql with where clause but after put secound value laravel give me a blank result? If i try to filtered with first value example like this : http://localhost/transport/1 everything is good but if i try to set from destionation give me a blank result. example with fail : http://localhost/transport/1/Германия
Here is my Controller
class TransportController extends Controller
{
public function filtermethod($method){
$data['ads'] = db::table('ads')->where('method', $method)->get();
return view('transport', $data );
}
public function regionfrom($from){
$data['ads'] = db::table('ads')->where('from', $from)->get();
return view('transport', $data );
}
Here is my routes :
Route::get('transport/{method}', 'TransportController#filtermethod');
Route::get('transport/{method}/{from}', 'TransportController#regionfrom');
Your second route should be giving your controller 2 variables.
public function regionfrom($method, $from)
Is what your route your having problems with is calling, do the logic you like in there.
If you would like to filter twice, try this:
$data = DB::table('ads')-where('method', $method)->where('region', $region)->get();
So I have a website where I buy items. I do an if to see if the session data "moedas" (user's coins) is greater than or equal to the item's price. I want to get the item's data from my model where I have the model's function dados_acessorio. So I call this function on my controller and what I want is to receive that data and be able to use it on the same controller after. The problem is I am getting an error. I am new to php/CodeIgniter and I could use some help.
Here is the controller's function:
public function confirmarCompra($id){
$this->load->model('acessorios_model','acessorios');
$dados['acessorio'] = $this->acessorios->dados_acessorio($id)[0];
if($this->session->moedas >= $acessorio['preco']){
}
}
Here is the model's function:
public function dados_acessorio($id){
return $this->db->from('acessorios')->where('id_acessorio', $id)->get()->result_array();
}
Here is an image of my acessorios table:
https://imgur.com/tth9mKS
The error that I receive on the controller:
https://imgur.com/BdR2ZMq
The error is on the following line:
if($this->session->moedas >= $acessorio['preco'])
Thanks!
Looks like it could be a typo? $acessorio is how it will appear in the view.
change this:
if($this->session->moedas >= $acessorio['preco']){
to this
if($this->session->moedas >= $dados['acessorio']['preco']){
I am trying to make my URL more SEO friendly on my Laravel application by replacing the ID number of a certain object by the name on the URL when going to that specific register show page. Anyone knows how?
This is what I got so far and it displays, as normal, the id as the last parameter of the URL:
web.php
Route::get('/job/show/{id}', ['as'=>'website.job.show','uses'=>'HomeController#show']);
Controller method
public function show($id){
$job = Job::findOrFail($id);
return view('website.job')->with(compact('job'));
}
Blade page where there is the link to that page
{{$job->name}}
You can overwrite the key name of your Job model:
public function getRouteKeyName()
{
return 'name';
}
Then in your route simply use {job}:
Route::get('/job/show/{job}', ...);
And to call your route:
route('website.job.show', $job);
So your a tag would look like this:
{{ $job->name }}
Inside your controller, you can change the method's signature to receive the Job automatically:
public function show(Job $job)
{
return view('website.job')
->with(compact('job'));
}
For more information, look at customizing the key name under implicit binding: https://laravel.com/docs/5.8/routing#implicit-binding
You need simply to replace the id by the name :
Route::get('/job/show/{name}', ['as'=>'website.job.show','uses'=>'HomeController#show']);
In the controller action:
public function show($name){
//Make sure to replace the 'name' string with the column name in your DB
$job = Job::where('name', $name)->first();
return view('website.job')->with(compact('job'));
}
Finally in the blade page :
{{$job->name}}
2 options:
1) one is like #zakaria-acharki wrote in his comment, by the name of the job and search by the name for fetching the data
2) the second is to do it like here in stackoverflow
to build the url with the id/name
in this way you will make sure to fetch and show the relevant job object by the unique ID
the route:
Route::get('/job/show/{id}/{name}', ['as'=>'website.job.show','uses'=>'HomeController#show']);
in the controller, update the check if the name is equal to the job name (in case it was changed) to prevent duplicate pages url's
public function show($id, $name){
$job = Job::findOrFail($id);
// check here if( $job->name != $name ) {
// redirect 301 to url with the new name
// }
return view('website.job')->with(compact('job'));
}
in the blade.php :
{{$job->name}}
I have this function in one of my controllers:
public function getApplicationFiles(Contract $contract) {
dd($contract->get());
}
The parameter is defind in the url. So my route is
Route::get('contracts/files/{contract}', 'ContractController#getApplicationFiles');
My problem is that the function getApplicationFiles displays all entries from the Contract type and not only the Object with the given id?
What am I doing wrong?
I would go with the following, make your route a named route like this:
Route::get('contracts/files/{contract}', 'ContractController#getApplicationFiles')->name('contract.files');
Then in your view use it like this:
Files
This should give you the expected result.
u don't need to pass all row u need to pass only one row try like this(if you are paassing only contract id in request)
public function getApplicationFiles(Contract $contract) {
//dd($contract);
return response()->json($contract); // if you pass json
return view('show',compact($contract); /// if you view
}
I need a little help and I can’t find an answer. I would like to replicate a row from one data table to another. My code is:
public function getClone($id) {
$item = Post::find($id);
$clone = $item->replicate();
unset($clone['name'],$clone['price']);
$data = json_decode($clone, true);
Order::create($data);
$orders = Order::orderBy('price', 'asc')->paginate(5);
return redirect ('/orders')->with('success', 'Success');
}
and i got an error :
"Missing argument 1 for
App\Http\Controllers\OrdersController::getClone()"
.
I have two models: Post and Order. After trying to walk around and write something like this:
public function getClone(Post $id) {
...
}
I got another error
Method replicate does not exist.
Where‘s my mistake? What wrong have i done? Maybe i should use another function? Do i need any additional file or code snippet used for json_decode ?
First of all, make sure your controller gets the $id parameter - you can read more about how routing works in Laravel here: https://laravel.com/docs/5.4/routing
Route::get('getClone/{id}','YourController#getClone');
Then, call the URL that contains the ID, e.g.:
localhost:8000/getClone/5
If you want to create an Order object based on a Post object, the following code will do the trick:
public function getClone($id) {
// find post with given ID
$post = Post::findOrFail($id);
// get all Post attributes
$data = $post->attributesToArray();
// remove name and price attributes
$data = array_except($data, ['name', 'price']);
// create new Order based on Post's data
$order = Order::create($data);
return redirect ('/orders')->with('success', 'Success');
}
By writing
public function getClone(Post $id)
you are telling the script that this function needs a variable $id from class Post, so you can rewrite this code like this :
public function getClone(){
$id = new Post;
}
However, in your case this does not make any sence, because you need and integer, from which you can find the required model.
To make things correct, you should look at your routes, because the url that executes this function is not correct, for example, if you have defined a route like this :
Route::get('getClone/{id}','YourController#getClone');
then the Url you are looking for is something like this :
localhost:8000/getClone/5
So that "5" is the actual ID of the post, and if its correct, then Post::find($id) will return the post and you will be able to replicate it, if not, it will return null and you will not be able to do so.
$item = Post::find($id);
if(!$item){
abort(404)
}
Using this will make a 404 page not found error, meaning that the ID is incorrect.