如何在laravel sql查询中获得灵活性

Question

I want to add a sql filter where('comment_id', '=', 1) to php code

$datas = $this->model->ADD HERE->orderBy('created_at', 'DESC')->paginate(15);

Trying to add the string to code take me hours. How to make it?

---

Here is my code:

CommentResource.php passing the sql filter as string parameter.

<?php
class CommentResource extends BaseResource
{    
    public function index()
    {
        $filter = "where('comment_id', '=', 1)";
        return parent::index_filter($filter);
    }

CommentResource.php

<?php
class BaseResource extends Controller
{
    protected function index_filter($filter)
    {
        $datas = $this->model->ADD HERE->orderBy('created_at', 'DESC')->paginate(15);
        return view($this->resourceView.'.index')->with('datas', $datas);
    }
}

Answer 1

As I understand you want to use different types of where as filters in your queries. That's why you want to make them dynamic. I would suggest the following solution for your task:

<?php
class CommentResource extends BaseResource
{    
    public function index()
    {
        $filter = [ 'operator' => 'where', 'args' => ['comment_id', '=', 1]];
        return parent::index_filter($filter);
    }

<?php
class BaseResource extends Controller
{
    protected function index_filter($filter)
    {
        $where = $filter['operator'];
        $args = $filter['args'];
        $datas = $this->model->$where(...$args)->orderBy('created_at', 'DESC')->paginate(15);
        return view($this->resourceView.'.index')->with('datas', $datas);
    }
}

However, it will work starting from Php5.6+ because of oeprator ...