Filament•14mo ago

Table tabs from database

I looked at the demo application and see the below tab on the table but instead of passing static value i want to retrieve fom the database. : public function getTabs(): array { return [ null => ListRecords\Tab::make('All'), 'new' => ListRecords\Tab::make()->query(fn ($query) => $query->where('status', 'new')), 'processing' => ListRecords\Tab::make()->query(fn ($query) => $query->where('status', 'processing')), 'shipped' => ListRecords\Tab::make()->query(fn ($query) => $query->where('status', 'shipped')), 'delivered' => ListRecords\Tab::make()->query(fn ($query) => $query->where('status', 'delivered')), 'cancelled' => ListRecords\Tab::make()->query(fn ($query) => $query->where('status', 'cancelled')), ]; } instead of new, processing, etc. i want to get a branch name

4 Replies

Kenneth Sese•14mo ago

@Bloom getTabs() just returns an array so you can do whatever you there. So you could do something like:

public function getTabs(): array
{
    $branches = Branch::all();

    $tabs = [];

    foreach ($branches as $branch) {
        $tabs = [
            $branch->name => Tab::make()->query(fn ($query) => $query->where('branch.name', $branch->name)),
        ];
    }

    return $tabs;
}

public function getTabs(): array
{
    $branches = Branch::all();

    $tabs = [];

    foreach ($branches as $branch) {
        $tabs = [
            $branch->name => Tab::make()->query(fn ($query) => $query->where('branch.name', $branch->name)),
        ];
    }

    return $tabs;
}

Adjust and refactor as needed of course.

BloomOP•14mo ago

Thank you it is working, but i want to return all the data also, can you help me with that

Kenneth Sese•14mo ago

Lots of ways but one way would be:

// $tabs = [];

$tabs = [
  null => Tab::make('All'),
];

// $tabs = [];

$tabs = [
  null => Tab::make('All'),
];

So instead of starting with an empty array, just start with your all tab.

BloomOP•14mo ago

Thank you it worked

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Table tabs from database