Is there a way to apply certain conditions to the list of records displayed by Voyager's BREAD? For example, only display records where a certain column is blank (i.e. WHERE 'col_name' IS NULL)?
UPDATE (17/12/18):
You can override the controller and add the where clause. For example, if you need to override "Posts" list, then you should:
Make a custom PostController: php artisan make:controller PostController
Extend it from Voyager BREAD controller: class PostController extends \TCG\Voyager\Http\Controllers\VoyagerBaseController
Import Voyager's Facade: use TCG\Voyager\Facades\Voyager;
Copy the whole index function from VoyagerBaseController. You can find it in \vendor\tcg\Voyager\Http\Controllers\VoyagerBaseController
Override it to add any logic or filter you need, for example:
$dataType = Voyager::model('DataType')
->where('slug', '=', $slug)
->where('col_name', '=', NULL)
->first();
You can also set $orderBy = 'col_name' and $sortOrder = 'asc'/'desc' custom values in the same function. Here Voyager's docs.
OLD:
Yes, there is. You need to edit the view and apply a condition filter. Here is explained how to override views (and controllers if you want to filter it before sending the data to the view).
Just define a Scope in your model:
public function scopeMyScope($query)
{
return $query->where('col_name', null);
}
then go to Bread Form and select it form Scope drop down :)
Related
I'm new to Laravel. I am developing a project. and in this project I have 4 tables related to each other
-Users
-Orders
-OrderParcels
-Situations
When listing the parcels of an order, I want to get the information of that order only once, the user information of that order once again, and list the parcels as a table under it. so far everything ok. but I also want to display the status of the parcels listed in the table as names. I couldn't add the 4th table to the query. do you have a suggestion? I'm putting pictures that explain the structure below.
My current working code is
$orderParcels = Orders::whereId($id)
->with('parcels')
->with('users:id,name')
->first();
and my 'orders' model has method
public function parcels(){
return $this->hasMany(OrderParcels::class);
}
public function users(){
return $this->hasOne(User::class,'id','affixer_id');
}
Note[edit]: I already know how to connect like this
$orderParcels = DB::table('order_parcels as op')
->leftjoin('orders as o','op.orders_id','o.id')
->leftjoin('users as u','o.affixer_id','u.id')
->leftjoin('situations as s','op.status','s.id')
->select('op.*','o.*','u.name','s.situations_name')
->where('op.orders_id',$id)->get();
but this is not working for me, for each parcels record it returns me orders and user info. I want once orders info and once user info.
Laravel provides an elegant way to manage relations between models. In your situation, the first step is to create all relations described in your schema :
1. Model Order
class User extends Model {
public function parcels()
{
return $this->hasMany(OrderParcels::class);
}
public function users()
{
return $this->hasOne(User::class,'id','affixer_id');
}
}
2. Model Parcel
class Parcel extends Model {
public function situations()
{
return $this->hasOne(Situation::class, ...);
}
}
Then, you can retrieve all desired informations simply like this :
// Retrieve all users of an order
$users = $order->users; // You get a Collection of User instances
// Retrieve all parcels of an order
$parcels = $order->parcels; // You get a Collection of User instances
// Retrieve the situation for a parcel
$situations = $parcel->situations // You get Situation instance
How it works ?
When you add a relation on your model, you can retrieve the result of this relation by using the property with the same name of the method. Laravel will automatically provide you those properties ! (e.g: parcels() method in your Order Model will generate $order->parcels property.
To finish, in this situation where you have nested relations (as describe in your schema), you should use with() method of your model to eager load all the nested relation of order model like this :
$orders = Orders::with(['users', 'parcels', 'parcels.situations'])->find($id)
I encourage you to read those stubs of Laravel documentation :
Define model relations
Eager loading
Laravel Collection
Good luck !
Use join to make a perfect relations between tables.
$output = Orders::join('users', 'users.id', '=', 'orders.user_id')
->join('order_parcels', 'order_parcels.id', '=', 'orders.parcel_id')
->join('situations', 'situation.id', '=', 'order_parcels.situation_id')
->select([
'orders.id AS order_id',
'users.id AS user_id',
'order.parcels.id AS parcel_id',
'and so on'
])
->where('some row', '=', 'some row or variable')->get();
I'm trying to return a table without deleted rows (softdeletes)
This is my code
public function getMailRecipients($meoId){
return DB::table('mail_recipients')->where('meo_id', '=', $meoId)->select('name', 'email')->get();
}
but I get all rows, even those removed through softdeletes, What else should I add to avoid that?
thanks!
you are using query builder (facade DB) in this case you should do this:
DB::table('mail_recipients')->where('meo_id', '=', $ meoId)->whereNull('deleted_at')->select('name', 'email')->get();
If you use the model, you must use the SoftDeletes trait
class Flight extends Model{
use SoftDeletes;
}
see more in the documentation
https://laravel.com/docs/8.x/eloquent#soft-deleting
Note *: The soft deleting feature works when using Eloquent. If you are querying the results with query builder you will eventually see all the records trashed and not trashed.
You can try with this:
public function getMailRecipients($meoId)
{
return DB::table('mail_recipients')
->whereNull('deleted_at')
->where('meo_id', $meoId)
->get(['name', 'email']);
}
If you arrived here because you are using protected $softDelete = true in your model and Laravel seems to ignore it, the solution is to use Illuminate\Database\Eloquent\SoftDeletes in your model. Works everytime!
I'm learning how to create web application with laravel and I don't know correct way to create struct for this example form:
I have simple controller named i.e. PostController with methods:
index(), create(), store(), show(), edit(), update(), destroy()
created via make:controller --resource.
And I have routes for it in my web.php file created via Route::resource() method.
Now, my create() method in PostController display a form for create new post, but this form required select an option from list.
I can't display whole list because it's thousands of options do I want create search field which use ajax to get available options from my database. Then user just select an option and can send form.
It's ofcourse just example but in that situation where should I put my method controller and route for search options?
Should I crate additional method in my PostController ( i.e. searchOption( $string ) ) or should it be new controller?
If it should be new controller then do you have any suggestion for folder/name structure for it?
And where should I put my routes for it? Still in web.php? or api.php?
And again, how should I name this route?
Sinse you have another entity you can make another controller. Let's say you want filtered City in your Post's create view.
// Controllers/Api/CityController.php
public function index(Request $request)
{
$query = City::query();
if ($request->has('search')) {
$query->where('name', 'LIKE', $request->input('search') . '%');
}
$cities = $query->paginate();
return response()->json(['cities' => $cities]);
}
I created BREAD for using voyager with relationship column to display all the users and i want this column to show only users with admin role.
I tried to override the view but I think its wrong choice any help?
#if($row->field == 'lead_belongsto_user_relationship')
#endif
I can reach the column in the view/add view but I don't know how to edit the array of results or how to override the query.
It is showing all the users I want to show only admin list.
I finally did it Using custom Controllers.
Steps:
First create new controller
php artisan make:controller LeadController
Extend Voyagers controller
<?php
namespace App\Http\Controllers;
class LeadController extends \TCG\Voyager\Http\Controllers\VoyagerBaseController
{
//...
}
?>
Finally you can override the relation function from VoyagerBaseController
if($request->type == "lead_belongsto_user_relationship"){
if ($search) {
$total_count = app($options->model)->where($options->label, 'LIKE', '%'.$search.'%')->count();
$relationshipOptions = app($options->model)->take($on_page)->skip($skip)
->where($options->label, 'LIKE', '%'.$search.'%')->where("role_id",3)
->get();
} else {
$total_count = app($options->model)->count();
$relationshipOptions = app($options->model)->take($on_page)->skip($skip)->where("role_id",3)->get();
}
}
Here i overrided it to select only users with role_id = 3 you can customize it as you need.
After that go to the BREAD-settings and fill in the Controller Name
with your fully-qualified class-name in my case it is
"\App\Http\Controllers\LeadController"
I am trying to retrieve the thumb image path by joining the images table to the listing table. As such, I have the following query in my controller.
$listings = Listing::select('listings.*, images.path as image_path')
->where('listings.ownerid', '=', $ownerid)
->leftJoin('images', 'listings.thumbId', '=', 'images.id')->get();
After testing out the function, the query fails since laravel interprets the query as
select `listings`.`*, images`.`path` as `image_path` from `listings` left join `images` on `listings`.`thumbId` = `images`.`id` where `listings`.`ownerid` = 1)
Notice the asterisk (*) is joined with the ", images" word making it '*, images'.
The query works fine without laravel's odd typo. How does one fix this issue?
You need to do one change in your query. You are passing raw select fields so you need to use selectRaw() instead of select(). Like
$listings = Listing::selectRaw('listings.*, images.path as image_path')
->where('listings.ownerid', '=', $ownerid)
->leftJoin('images', 'listings.thumbId', '=', 'images.id')->get();
check by try above query.
I suggest you to use Laravel Eloquent Relationships feature. Since your code above is more like Query Builder rather than Eloquent. Let's see the example bellow:
You will have 2 Models, 1 for each table (listings, images):
App\Listing Model:
<?php
...
use App\Image;
class Listing extends Eloquent {
...
protected $table = 'listings';
// define Eloquent Relationship of Listing model to Image model
public function image() {
return $this->belongsTo(Image::class, 'thumbId');
}
...
}
App\Image Model:
<?php
...
use App\Listing;
class Image extends Eloquent {
...
protected $table = 'images';
...
// define Eloquent Relationship of Image model to Listing model
public function listings() {
return $this->hasMany(Listing::class, 'thumbId');
}
}
So how to get the data?
// get all listing data
$listings = Listing::all();
// loop through the data
foreach ($listing as $listing) {
dump($listing->id);
// because we have define the relationship, we can access the related data of image
dump($listing->image->path);
// call $this->image will return related Image model
dump($listing->image);
}
You can see Laravel official documentation for more example and explanation.
Hope it helps.