Fouladgar的简单使用Eloquent Builder class return 错误"it is not an interface"

Question

我在创建使用 Eloquent 构建器 class 的简单过滤器时遇到问题。 Returns我的错误

App\EloquentFilters\Property\RoomsFilter cannot implement Fouladgar\EloquentBuilder\Support\Foundation\Contracts\Filter - it is not an interface

基本上是从如何开始复制粘贴的

<?php
namespace App\EloquentFilters\Property;
use Fouladgar\EloquentBuilder\Support\Foundation\Contracts\Filter;
use Illuminate\Database\Eloquent\Builder;
class RoomsFilter implements Filter
{
    public function apply(Builder $builder, $value): Builder
    {
        return $builder->where('rooms', $value);
    }
}

感谢大家的帮助

Answer 1

您需要扩展它，而不是像错误所说的那样实施它。

变化：

 class RoomsFilter implements Filter
{ }

到

class RoomsFilter extends Filter
{ }

来自gitHub repo：

Writing a filter is simple. Define a class that extends the Fouladgar\EloquentBuilder\Support\Foundation\Contracts\Filter abstract class. This class requires you to implement one method: apply.

Fouladgar的简单使用Eloquent Builder class return 错误"it is not an interface"

Simple use of Fouladgar's Eloquent Builder class return error "it is not an interface"

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