i want to filter a collection in order to return only items with fullname that is like the one given in parmaters
public function searchFriend($fullName){
return Auth::user()->friends->filter(function ($item) use ($fullName) {
return false !== stristr($item->friend->full_name, $fullName);
});
}
the function actually is not returning the correct results.
Instead of accessing the collection directly, you can do the filtering in SQL.
public function searchFriend($fullName){
return Auth::user()
->friends()
->where('full_name', 'like', '%'.$fullName.'%')
->get();
}
If you, for whichever reason, need to do it on the collection, then the problem with your current code is that $item represents the friend, so you're checking $item->friend->full_name instead of $item->full_name.
public function searchFriend($fullName){
return Auth::user()
->friends
->filter(function ($item) use ($fullName) {
return false !== stristr($item->full_name, $fullName);
});
}
I guess friend is a relationship to the User model so it could be
public function searchFriend($fullName)
{
return Auth::user()->friends()->whereHas('friend', function ($query) use ($fullName) {
$query->where('full_name', 'like', "%{$fullName}%");
})->get();
}
If you have first_name and last_name in User model then you can use concat.
public function searchFriend($fullName)
{
return Auth::user()->friends()->whereHas('friend', function ($query) use ($fullName) {
$query->whereRaw("concat(first_name, ' ', last_name) like ?", ["%{$fullName}%"]);
})->get();
}
Related
I have that project that have orders that have status, and every status can be seen depending on if the user have the permission to manage it. I've tried to use custom collection but the downside is it's not paginatiable by default,
here's the code
class OrdersCollection extends Collection
{
public function allowedForUser(User $user)
{
return $this->filter(function ($order) use ($user) {
return $user->can(sprintf('manage %s orders', strtolower($order->status)));
});
}
}
here's how I'm trying to use it
$orders = Order::withoutTrashed()
->allowedForUser(Auth::user())
->paginate(50);
EDIT:
to avoid filtering through 20k+ rows in the table I came up with this custom builder
class OrderBuilder extends Builder
{
public function WhereAllowedForUser(User $user)
{
$permissions = $user->permissions()
->where('name', 'like', 'manage % orders')
->pluck('name')
->map(function ($item) {
return str_replace('manage ', '', str_replace(' orders', '', $item));
});
return $this->whereIn('status', $permissions);
}
}
You have to use scoped queries for your Model.
The idea is for you to pass a value to the scope and do what you want/need with it:
class Order extends Model
{
public function scopeWhereAllowedForUser($query, User $user)
{
$permissions = $user->permissions()
->where('name', 'like', 'manage % orders')
->pluck('name')
->map(function ($item) {
return str_replace('manage ', '', str_replace(' orders', '', $item));
});
return $query->whereIn('status', $permissions);
}
}
So you use it like this:
$orders = Order::whereAllowedForUser(Auth::user())->paginate(50);
I am not 100% sure what you are trying to do with that $permissions query, I am not sure what you tried to do, so explain it a little more and I can help you with it (better code).
Have a query, how I can filter results by translation relation (by name column)
$item = Cart::select('product_id','quantity')
->with(['product.translation:product_id,name','product.manufacturer:id,name'])
->where($cartWhere)
->get();
my model
Cart.php
public function product($language = null)
{
return $this->hasOne('App\Models\Product','id','product_id');
}
Product.php
public function translations()
{
return $this->hasMany('App\Models\ProductTranslation','product_id','id');
}
Update v1.0
do like this, but query takes too long time
$item = Cart::select('product_id','quantity')
->with(['product.translation', 'product.manufacturer:id,name'])
->where($cartWhere)
->when($search,function ($q) use ($search) {
$q->whereHas('product.translation', function (Builder $query) use ($search) {
$query->where('name', 'like', '%'.$search.'%');
$query->select('name');
});
}
)
->get() ;
Inside the array within your with() method, you can pass a function as a value.
Cart::select('product_id','quantity')
->with([
'product', function($query) {
$query->where($filteringAndConditionsHere);
}
]);
https://laravel.com/docs/7.x/eloquent-relationships#eager-loading
For example I can use:
$address = $user->address
which will return the addresses for the user.
// User.php (model)
public function address()
{
return $this->hasMany(Address::class, 'refer_id', 'id');
}
public function billingAddress()
{
return $this->address()
->where('type', '=', 1)
->where('refer_id', '=', $this->id)
->first();
}
However, I would like to return the BillingAddress for the user depending on this where clause. How do I do it?
EDIT:
If I use this inside... OrderController#index it returns correctly
$orders = Order::with('order_fulfillment', 'cart.product', 'user.address', 'payment')->get();
return new OrderResource($orders);
However, If I change it to:
$orders = Order::with('order_fulfillment', 'cart.product', 'user.billingAddress', 'payment')->get();
return new OrderResource($orders);
I get this error:
Symfony\Component\Debug\Exception\FatalThrowableError
Call to a member function addEagerConstraints() on null
One option is you can use whereHas in your query. For example,
$orders = Order::with('order_fulfillment', 'cart.product', 'user.address', 'payment')
->whereHas(
'address', function ($query) {
$query->where('type', '=', 1)
->first();
}
)->get();
return new OrderResource($orders);
This is one option. try to dd($orders) an find if its working.
You had an another option like this, in your model
public function address()
{
return $this->hasMany(Address::class, 'refer_id', 'id');
}
Add relations like
public function billingAddress()
{
return $this->hasOne(Address::class, 'refer_id', 'id')->where('type', 1);
}
And
public function shippingAddress()
{
return $this->hasOne(Address::class, 'refer_id', 'id')->where('type', 2);
}
Then in your query,
$orders = Order::with('order_fulfillment', 'cart.product', 'user.address','user.billingAddress', 'user.shippingAddress', 'payment')->get(); return new OrderResource($orders);
I have this DB structure:
classroom_user is Elqouent many-to-many pivot.
User.php:
public function classrooms() {
return $this->belongsToMany(ClassRoom::class, 'classroom_user', 'classroom_id', 'user_id');
}
ClassRoom.php:
public function users()
{
return $this->belongsToMany(User::class, 'classroom_user', 'user_id', 'classroom_id');
}
I have a $user and want to write Eloquent method which gets all users who are in at least one of classrooms the $useralso is in. Something like $user->classrooms->users. But this falls with error Property [users] does not exist on this collection instance.. How can I do this?
I think you can do it like this:
$users = User::whereHas('classrooms', function($query) {
$query->whereHas('users', function($query) {
$query->where('id', auth()->user()->id);
});
})->get();
You could store this as a scope in your User model:
public function scopeClassmatesOf($query, $user)
{
$query->whereHas('classrooms', function($query) use($user) {
$query->whereHas('users', function($query) use($user) {
$query->where('id', $user->id);
});
})
}
And then call it from a controller like:
$user = auth()->user();
$users = User::classmatesOf($user)->get();
Your asking for users of a collection. $user->classrooms is a collection of classrooms.
If you do $user->classrooms->first()->users you will get all the users from the first classroom.
To get all users from all classrooms, you can do something like;
$users = $user->classrooms->map(function ($classroom) {
return $classroom->users;
});
Note that will return duplicate users and authenticated user him self witin the collection. So, you might need to do some filtering to the result.
This worked for me:
public function classmates() {
$self = $this;
$a = User::with('classrooms')->whereHas('classrooms', function ($query) use ($self) {
$query->whereHas('users', function ($query) use ($self) {
$query->where('users.id', $self->id);
});
});
return $a;
}
I have four models: OwnerCompany, Owner, User, and Role.
I want to get all OwnerCompanys eager loading their Owner and also eager loading its Users who have the Role with the name 'admin'.
OwnerCompany::with('owner.users')->whereHas('owner.users.roles', function ($query) {
$query->where('name', 'admin');
})->get();
This loads in the models but doesn't constrain the users, they are all loaded.
Not tested . I think you are missing like or equal operator.
OwnerCompany::with('owner.users')->whereHas('owner.users.roles', function ($query) {
$query->where('name','like','admin');
})->get();
I try this, if work then let me know. not tested
1).
OwnerCompany::with('owner.users.roles')->whereHas('owner', function ($query) {
$query->whereHas('users', function($qr){
$qr->whereHas('roles', function($q){
$q->where('name', 'admin');
});
});
})->get();
OR
2).
OwnerCompany::with('owner.users.roles')->whereHas('owner.users.roles', function ($query) {
$query->where('name', 'admin');
})->get();
If I got it right, you want each OwnerCompany object to contain the Owner object, which will contain "admins". I would do something like this.
$ownerCompany = OwnerCompany::all();
return $ownerCompany->each(function ($company) {
$company->owner;
$company->owner->users->each(function ($user) {
return $user->hasRole('admin');
});
});
and inside User class:
public function role()
{
return $this->belongsTo(Role::class);
}
public function hasRole($role)
{
if ($this->role()->where('name', $role)->first()) {
return $this;
}
return null;
}
Unfortunately, I can't test it right now, so If you have any trouble let me know.