php - 尝试在 Codeigniter 4 中启动分页时返回错误

Question

我正在尝试按照 documentation 在 Codeigniter 4 中启动分页功能.

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

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

我的 Controller 代码如下:

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");

}

但是，在运行它时，我收到以下错误:
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 44 7

这是我的模型

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;
    }
}

任何解决此问题的帮助将不胜感激。

Answer 1

我不知道你到底在哪里得到这个错误，但我在你的代码中发现了一些错误。尝试修复这些错误，也许对您有所帮助。以下是错误:

paginate()方法返回结果，因此它必须是链中的最后一个。
示例:$this->jobs->all_jobs()->paginate(20)

您可以获得 Pager像这样:$this->jobs->pager

如果您想使用您的 all_jobs()方法与 paginate()方法，你必须在你的 all_jobs() 中返回一个模型方法

这是您的 Controller 的正确代码:

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");
}

这是您模型的正确代码:

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;
}

要查看导致错误的确切位置，请尝试设置 CI_ENVIRONMENT = development在您的 .env根目录下的文件。之后尝试重新加载发现此错误的页面。您将看到带有回溯的 CodeIgniter 错误页面。尝试从回溯中复制所有数据并将其放在这里，这有助于了解到底发生了什么。