Error returned when trying to initiate pagination in Codeigniter 4

Question

I'm trying to initiate the pagination function in Codeigniter 4 as per the documentation.

$model = new \App\Models\UserModel();

$data = [
    'users' => $model->paginate(10),
    'pager' => $model->pager
];

My controller code is the following:

public function jobmarket() {
    $this->jobs = new \App\Models\Jobs();

    if (!$this->ionAuth->loggedIn())
    {
        return redirect()->to('/logg-inn');
    }

    echo view("dashboard/header", ([
        'ionAuth' => $this->ionAuth,
        'uri' => $this->uri,
    ]));

    echo view("dashboard/jobmarket", ([
        'session' => $this->session,
        'ionAuth' => $this->ionAuth,
        'validation' => $this->validator,
        'jobs' => $this->jobs->paginate(20)->all_jobs(),
        'pager' => $this->jobs->pager()->all_jobs(),
    ]));

    echo view("assets/footer");

}

However, when running this I get the following error:

Argument 1 passed to CodeIgniter\Database\BaseResult::getResult() must be of the type string, null given, called in xxxx/app/vendor/codeigniter4/framework/system/Model.php on line 447

This is my model

public function all_jobs() {
    $this->categories = new \App\Models\Categories();
    $builder = $this->db->table('jobs');
    $builder->select('*');
    $builder->join('users', 'users.id = jobs.jobs_u_id');
    $builder->join('categories', 'category_id = jobs.jobs_category');
    // Make sure to not show current user's listings, since these will show up under "My listings"
    $builder->where('jobs_u_id !=', $this->current_user->id);
    // Check that the listing reflects users chosen categories
    $builder->whereIn('category_id', $this->categories->user_categories());
    $builder->orderBy('jobs_id', 'desc');
    $query =  $builder->get();
    if ($builder->countAllResults() > 0)
    {
        return $query->getResult();

    } else {
        return false;
    }
}

Any help to resolve this would be appreciated.

Can you try without pagination? Seems the error is regarding the Model and Not pagination itself. Also, your pager implementation may not be correct. I tried the $model->pager but it wasn't working for me so then I changed it to $inventoryModel->pager->getDetails(). Try changing $this->jobs->pager()->all_jobs() to $this->jobs->pager->getDetails() and see if it works out. — Dhaval Chheda, Apr 13 '20 at 18:24
Also, checkout the pager class from the code. It will give you some idea. But most probably I think your error is related to the model itself. — Dhaval Chheda, Apr 13 '20 at 18:26
$builder->join('categories', 'category_id = jobs.jobs_category'); Shouldn't it be categories.id or categories.category_id. I suspect category_id is from the users table. Comment out joins and ad them one by one. To see if sql is right print_r($this->db->last_query()); print_r($query) to see result. Run the query in phpmyadmin or mysql developer to see what it returns. — Peter Szalay, Apr 28 '20 at 20:49

Oleg Demkiv · Answer 1 · 2020-04-24T18:53:39.623

I don't know where you exactly get this error, but I found a few bugs in your code. Try to fix these bugs, maybe it helps you. Here are the bugs:

The paginate() method returns the result, so it must be the last in your chain. Example: $this->jobs->all_jobs()->paginate(20)
You can get a Pager like this: $this->jobs->pager
If you want to use your all_jobs() method with the paginate() method, you have to return a model in your all_jobs() method

Here is the correct code for your controller:

public function jobmarket() {
    $this->jobs = new \App\Models\Jobs();

    if (!$this->ionAuth->loggedIn())
    {
        return redirect()->to('/logg-inn');
    }

    echo view("dashboard/header", ([
        'ionAuth' => $this->ionAuth,
        'uri' => $this->uri,
    ]));

    echo view("dashboard/jobmarket", ([
        'session' => $this->session,
        'ionAuth' => $this->ionAuth,
        'validation' => $this->validator,
        'jobs' => $this->jobs->all_jobs()->paginate(20),
        'pager' => $this->jobs->pager,
    ]));

    echo view("assets/footer");
}

And here is the correct code for your model:

public function all_jobs() {
    $this->categories = new \App\Models\Categories();
    $builder = $this->db->table('jobs');
    $builder->select('*');
    $builder->join('users', 'users.id = jobs.jobs_u_id');
    $builder->join('categories', 'category_id = jobs.jobs_category');
    // Make sure to not show current user's listings, since these will show up under "My listings"
    $builder->where('jobs_u_id !=', $this->current_user->id);
    // Check that the listing reflects users chosen categories
    $builder->whereIn('category_id', $this->categories->user_categories());
    $builder->orderBy('jobs_id', 'desc');

    return $this;
}

To see the exact place that causes the error try to set CI_ENVIRONMENT = development in your .env file in your root directory. After that try to reload the page where you found this error. You will see a CodeIgniter`s error page with a backtrace. Try to copy all data from backtrace and place here, it helps to understand what exactly is going on.

score 0 · Answer 2 · answered Jul 07 '20 at 01:23

In my case, I had to override findAll() with my own findAll() method, and also check that my specific model extended the Model class of CodeIgniter.

I was basically getting the same error as you @kanarifugl.

So I had something like this:

<?php namespace App\Models;

use CodeIgniter\Model;

class BlogModel extends Model
{
    protected $db;

    public function __construct() 
    {
        parent::__construct();
        $this->db = \Config\Database::connect();
        $this->table = 'Blog';
    }

    public function findAll(int $limit = 12, int $offset = 0)
    {
        $builder = $this->builder();
        $builder->select('*');
        $builder->orderBy('blog_id', 'DESC');
        $builder->join('Categories', 'Categories.category_id = Blog.category_id');

        if ($this->tempUseSoftDeletes === true)
        {
            $builder->where($this->table . '.' . $this->deletedField, null);
        }

        $row = $builder->limit($limit, $offset)
                ->get();

        $row = $row->getResult($this->tempReturnType);

        $eventData = $this->trigger('afterFind', ['data' => $row, 'limit' => $limit, 'offset' => $offset]);

        $this->tempReturnType     = $this->returnType;
        $this->tempUseSoftDeletes = $this->useSoftDeletes;

        return $eventData['data'];
    }
}

And then in my Blog controller I just had this:

<?php namespace App\Controllers;

class Blog extends BaseController
{
    private $blogModel;

    public function __construct() 
    {
        $this->blogModel = new \App\Models\BlogModel();
    }

    public function index()
    {
        $data = [
            'section' => 'blog',
            'articles' => $this->blogModel->paginate(12, 'blog'),
            'pager' => $this->blogModel->pager
        ];
        echo view('articules', $data);
    }
}

And to finish it off, in my view, the part that uses the pager was like this:

<?= $pager->links('blog') ?>

I hope this helps someone.

score 0 · Answer 3 · answered Jun 08 '21 at 16:38

0

Remove the constructor function from the Model class to solve the problem

> Argument 1 passed to CodeIgniter\Database\BaseResult::getResult()
> pagination

answered Jun 08 '21 at 16:38

Muhammad Azeem Abbas

21
3

Error returned when trying to initiate pagination in Codeigniter 4

3 Answers3