paginate

问题 Struggling to wrap my head around paginate in CI4. All examples in the user guide shows the ->paginate() method linked directly onto the $model variable (without arguments) in the Controller. I have all my queries in the Model, so I'm a bit confused if the paginate method goes onto the end of my query there, or where it's called in the Controller. I've tried both - neither work. Model: public function brand_name($brand_name_slug) { return $this ->db ->table('shop a') ->select() ->join('

问题 I have a merge on two query results which works fine: $events1 = \App\Event::Where('valid_to','>=',$today)->orderByRaw('valid_to','ASC')->get(); $events2 = \App\Event::Where('valid_to','<',$today)>orderByRaw('valid_to','ASC')->get(); $events = $events1->merge($events2); Now I need to paginate this new collection and as suggestted I added this piece: $page = 1; $perPage = 60; $pagination = new \Illuminate\Pagination\LengthAwarePaginator( $events->forPage($page, $perPage), $events->count(),

问题 I have a merge on two query results which works fine: $events1 = \App\Event::Where('valid_to','>=',$today)->orderByRaw('valid_to','ASC')->get(); $events2 = \App\Event::Where('valid_to','<',$today)>orderByRaw('valid_to','ASC')->get(); $events = $events1->merge($events2); Now I need to paginate this new collection and as suggestted I added this piece: $page = 1; $perPage = 60; $pagination = new \Illuminate\Pagination\LengthAwarePaginator( $events->forPage($page, $perPage), $events->count(),

问题 I import an old database (hyperfile) to mysql to create a newapp with Laravel. I'm using Laravel built in paginate method, but the first query "select count(*)" takes about 10 seconds to answer because there's about 70000 lines in this table. I don't really need to know how many pages are present...just two buttons "previous/next" should be enough. Is there a method to avoid this query ? 回答1: https://laravel.com/docs/5.2/pagination "Simple Pagination" If you only need to display simple "Next"

问题 I import an old database (hyperfile) to mysql to create a newapp with Laravel. I'm using Laravel built in paginate method, but the first query "select count(*)" takes about 10 seconds to answer because there's about 70000 lines in this table. I don't really need to know how many pages are present...just two buttons "previous/next" should be enough. Is there a method to avoid this query ? 回答1: https://laravel.com/docs/5.2/pagination "Simple Pagination" If you only need to display simple "Next"

问题 Is it possible to paginate and display number of pages in Coldfusion using only one query? My understanding is you can obviously paginate with one query, but you would need an additional query to create the pages. This is in order to calculate the total number of results. (currentPage - 1) * resultsPerPage = Offset in MySQL query. This logic is sufficient to create next/prev buttons. But in order to know the number of pages, would we not need to know the total number of results using a

问题 I am trying to make custom query integrated with a plugin to sort posts. Here's the code. index.php $view =$_GET["sort"]; if($view == "views"){ if ( have_posts() ) : $paged = ( get_query_var('paged') ) ? get_query_var('paged') : 1; $args = array('post_type' => 'post', 'meta_key' => '_kksr_avg', 'orderby' => 'meta_value', 'order' => 'desc', 'paged' => $paged, 'posts_per_page' => 5); $query = new WP_Query($args); while ( $query->have_posts() ) : $query->the_post(); // code for display posts

问题 How can I use paginate function from Eloquent in Slim 3 project using twig ? This is in my controller : $posts = Sound::paginate(2); $this->container->view->render($response, 'admin/sounds/index.twig', [ 'posts' => $posts ]); This is the view : {{ posts.links() }} But it doesn't work as well as I expected : Warning: call_user_func() expects parameter 1 to be a valid callback, no array or string given in **PATH_TO_PROJECT**\vendor\illuminate\pagination\AbstractPaginator.php on line 412 Fatal

问题 I'm using cakephp and would like to display all Submissions which are part of Category 'X' I have 4 tables with a HABTM relationship. Users -> (haveMany) -> Submissions <-> (hasAndBelongsToMany) <-> Categories but I would like to do so using the $this->paginate() and for each submission I would like to display the user who posted the submission. User Table Id | Name -----+------------------- 1 | User 1 2 | User 2 Submission Table Id | Name | User_id -----+-------------------+-------------- 1

问题 I'm using cakephp and would like to display all Submissions which are part of Category 'X' I have 4 tables with a HABTM relationship. Users -> (haveMany) -> Submissions <-> (hasAndBelongsToMany) <-> Categories but I would like to do so using the $this->paginate() and for each submission I would like to display the user who posted the submission. User Table Id | Name -----+------------------- 1 | User 1 2 | User 2 Submission Table Id | Name | User_id -----+-------------------+-------------- 1