Level 60
Polymorphic Relationships https://laravel.com/docs/11.x/eloquent-relationships#polymorphic-relationships
Or try union, should be very ease while the tables' structure is the same https://laravel.com/docs/11.x/queries#unions
2 likes
Jan 26, 2025
5
Level 1
List multiple similar relationships in a single table
Hello friends
I am developing a project with filament. There is an issue I am stuck.
I have relationships named cashMovements, bankMovements and bondMovements in my Customer model;
public function cashMovements()
{
return $this->hasMany(CustomerCashMovement::class, 'customer_account_id');
}
public function bankMovements()
{
return $this->hasMany(CustomerBankMovement::class, 'customer_account_id');
}
public function bondMovements()
{
return $this->hasMany(CustomerBondMovement::class, 'customer_account_id');
}
The database of these relationships is exactly the same. For 3 relationships I need to set up a separate relationship manager but this is not enough. The 3 relationships need to be listed by date in a single table. My purpose here is to show the customer's cash, bank and bond movements financially.
But somehow I could not combine these two in a single table.
Is there a way to do this? In addition, I think this merge need will be 5 pieces in the future.
Level 1
Thank you, I actually tried with union, but there was a column mismatch, so I got an error.
$cashMovements = $this->financeCashMovements()
->select(
'id',
'created_at',
'price',
'movement_type',
'notes',
DB::raw(“‘CASH’ as movement_model”)
)
->orderBy('created_at', 'asc');
$bankMovements = $this->financeBankMovements()
->select(
'id',
'created_at',
'price',
'movement_type',
'notes',
DB::raw(“‘BANK’ as movement_model”)
)
->orderBy('created_at', 'asc');
$bondMovements = $this->financeBondMovements()
->select(
'customer_finance_bond_movements.id',
'customer_finance_bond_movements.created_at',
'price',
'customer_finance_bonds.bond_movement_type as movement_type',
'customer_finance_bonds.name as notes',
DB::raw(“‘BOND’ as movement_model”)
)
->orderBy('created_at', 'asc');
$movements = $cashMovements->union($bankMovements)- >union($bondMovements);
This is how I put it together. I have an additional question. How can I show the price column in a column by summing it with the previous row in each row? For example; Line 1: price -150, balance 0 Line 2: price 100, balance -50 (100-150) Line 3: price 200, balance 150 (200-50)
How do you think I can do this without breaking the relation structure?
Level 1
@Sergiu17 Thank you so much, you made my day.
Level 60
@rmznatly simplest way is to store a reference to a previous record somewhere.
@php
$previousMovement = null
@endphp
@foreach($movements as $movement)
{{ $movement->balance += $previousMovement?->balance ?? 0 }}
{{ $movement->created_at }}
// at the end of the loop, set $previousMovement
@php
$previousMovement = $movement
@endphp
@endforeach
Level 1
Thanks for your answer, but the problem is that the relationship manager only accepts relationship data. The manual definitions I make outside the query are answered with “must be relation only” error.
For example ->union()->get() returns the following error. Filament\Resources\RelationManagers\RelationManager::getRelationship(): Return value must be of type Illuminate\Database\Eloquent\Relations\Relation|Illuminate\Database\Eloquent\Builder, Illuminate\Database\Eloquent\Collection returned
so I couldn't find a way to take action here. So if there is a way to return this to the relation manager after php operations, I think this problem is solved.
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